Download HKDSE Mathematics Indefinite Integration By Leon Lee 1

HKDSE Mathematics
Indefinite Integration
By Leon Lee
1. Introduction
Integration, either definite or indefinite, is a large topic in HKDSE Extended Module 2. With the
introduction of integrations by substitution and by parts, which were not required in HKCEE
Additional Mathematics, to the syllabus, a much wider variety of questions can be set compared
with the old syllabus. Therefore, students should pay much more attention to this topic, and try to
master this topic as well as possible for good results in M2 examination.
This note will demonstrate the techniques in solving problems involving indefinite integration as
detailed as possible. It will start from the basic problems, and gradually to the hardest problems
which involve advanced techniques in integration.
2. What is Indefinite Integration?
→ Indefinite integration can be considered the ‘reverse’ process of differentiation.
→ In differentiation, we find the derivative of a function.
In indefinite integration, we find the primitive function of a function.
→ In simpler words, if the derivative of F (x) is f (x ) , then F (x) is a primitive function of f (x ) ,
i.e.
d
F ( x) = f ( x) .
dx
→ The primitive function of a function is not unique in nature.
d
d
d
F ( x ) = [ F ( x ) + 1] = [ F ( x ) + 2] = f ( x) .
dx
dx
dx
Then F (x ) , F ( x) + 1 and F ( x) + 2 are all primitive functions of f (x ) .
Note that
When we replace 1 or 2 by any other real constants, say π , e , 2 or − 10 , we still get the
same result.
→ From the above results, we see that there are infinitely many primitive functions for any
integrable function.
For convenience, if
d
F ( x ) = f ( x ) , then we write
dx
∫ f ( x)dx = F ( x) + C .
differentiation
→ The constant C is called the constant of integration, and it is
arbitrary in nature.
→ The sign
∫
is called the integral sign, and f (x ) is called
F(x)+C
f (x)
the integrand.
indefinite integration
Quick Example:
As
d  x
1

=
, we have
dx  2  4 x
∫4
1
x
dx =
x
+C.
2
1
Figure 1 A diagrammatic representation of
differentiation and indefinite integration
3. Methods of Indefinite Integration
The methods of indefinite integration will be introduced below.
3.1. Elementary Integration
The following shows some fundamental indefinite integration results, which can be obtained
directly from differentiation.
(1)
∫ kdx = kx + C , where k is a constant
(2)
x n+1
n
x
dx
=
+ C , where n ≠ −1
∫
n +1
(3)
∫ x dx = ln x + C , where x ≠ 0
Important!
is the absolute sign. It is defined as:
1
 x , when x ≥ 0
x =
− x , when x < 0
e ax
+ C , where a ≠ 0
a
From (4), when a = 1 , we have
(4)
ax
∫ e dx =
(5)
∫e
x
dx = e x + C .
The following gives the proof of (3). The remaining is left to readers as an exercise.
Proof:
When x > 0 , ln x = ln x . We have
d
1
(ln x ) = .
dx
x
When x < 0 , ln x = ln(− x) . We have
d
1
1
.
[ln(− x )] =
( −1) =
dx
x
−x
1
Combining the results, we have ∫ dx = ln x + C .
x
Furthermore, we have the following properties. They can also be easily proved by employing
differentiation.
(6)
∫ kf ( x)dx = k ∫ f ( x)dx, where k is a constant
(7)
∫ [ f ( x) ± g ( x)]dx = ∫ f ( x)dx ± ∫ g ( x)dx
Using (7) repeatedly, we have
(8)
∫ [ f ( x) ± f
1
2
( x ) ± ... ± f n ( x ) ]dx = ∫ f 1 ( x ) dx ± ∫ f 2 ( x) dx ± ... ± ∫ f n ( x ) dx
2
Example 1
4
1

Find ∫  x 2 −  dx . (HKDSE Sample Paper)
x

Solution:
4
2
3
4
 2 4
1
1
1  1 
 2 1
4
2 3
4
2 2
4
2 
∫  x − x  dx = ∫ ( x ) + C1 ( x )  − x  + C2 ( x )  − x  + C3 ( x ) − x  +  − x  dx


4 1 

= ∫  x 8 − 4 x 5 + 6 x 2 − + 4 dx
x x 

4
= ∫ ( x 8 − 4 x 5 + 6 x 2 − + x −4 ) dx
x
9
6
3
x
x
x
x −3
=
− 4⋅
+ 6 ⋅ − 4 ln x +
+C
−3
9
6
3
x 9 2x 6
1
=
−
+ 2 x 3 − 4 ln x − 3 + C
9
3
3x
Note: Always add the constant C after indefinite integration!
To make it simple, you should add it whenever the integral sign
(Binomial Theorem)
∫
disappears.
Example 2
Find (a) ∫
(
)
2
x + 1 dx (b) ∫ 10 x dx .
Solution:
(a)
∫(
)
2
x + 1 dx
= ∫ ( x + 2 x + 1)dx
1
= ∫ ( x + 2 x 2 + 1) dx
3
x2
2
=
+ 2⋅ x2 + x + C
2
3
3
2
x
4
=
+ x2 + x +C
2 3
(b) We first let 10 x = e ax , where a is a constant.
Solving, we have a = ln 10 .
ln 10 x = ln e ax
∫ 10 dx
= ∫e
x ln 10 = ax
a = ln 10
x
x ln 10
dx
e x ln 10
+C
ln 10
10 x
=
+C
ln 10
=
3
3.2. Integration by Substitution
The above formulae are very limited in usage and cannot deal with most integrals like
2x
x +1
∫ x − 1dx or ∫ x 2 +1dx . We need a new method called ‘integration by substitution’ to deal
with these integrals.
Let u = g ( x ) . Then, we have
∫ f ( g ( x)) g ' ( x)dx = ∫ f (u )du .
The proof is given in the appendix of this note on p.46.
This method can be regarded as the ‘reverse’ of the chain rule in differentiation.
Refer to the following illustration to see how we can apply this method to find integrals:
2x
dx . We take u = g ( x ) = x 2 + 1 . Then, we have g' ( x ) = 2 x .
Consider ∫
2
x +1
∫
2x
2
x +1
1
⋅ 2 xdx
g ( x)
dx = ∫
1
g ' ( x )dx
g ( x)
=∫
1
2
= ∫ u du
−
Here, f ( x ) =
1
1
and g ( x) = x 2 + 1 .
x
= 2u 2 + C
= 2 x2 +1 + C
As seen from the above example, we can see that we express the integral in the form
∫ f ( g ( x)) g ' ( x)dx
and transform it into the form
∫ f (u )du for simpler calculation.
However, we usually use the following way:
Let u = x 2 + 1 , then du = 2 xdx , i.e. dx =
∫
2x
x2 +1
dx = ∫
2x
1
du .
2x
We express dx in
terms of du.
1
du
u 2x
⋅
1
2
= ∫ u du
−
1
2
= 2u + C
= 2 x2 +1 + C
We use a suitable substitution u = g (x ) at the beginning. We differentiate it with respect to x
(w.r.t. x, in short) to obtain du = g ' ( x )dx . By rearranging terms, we get dx =
1
du . We put
g ' ( x)
it into the original integral and express the whole integral in terms of the new variable u only.
Finally, we find the integral, and express the result in terms of the original variable x.
4
Example 3 (2012 DSE)
(a) Find
∫
x +1
dx .
x
(b) Using the substitution u = x 2 − 1 , find ∫
x3
dx .
x2 −1
Solution:
(a)
∫
x +1
1
dx = ∫ (1 + )dx = x + ln x + C
x
x
(b) Let u = x 2 − 1 , then du = 2 xdx , i.e. dx =
1
du .
2x
x3
1 u +1
∫ x 2 − 1 dx = 2 ∫ u du
1
1
(by (a))
= u + ln u + C
2
2
1
1
= ( x 2 − 1) + ln x 2 − 1 + C
2
2
1
1
1
= x 2 + ln x 2 − 1 + C ' , where C' = C −
2
2
2
Always remember:
Express your final answer
in terms of the original
variable!
*Think About*:
(1) Why did we not to include −
1
in final answer and use a new constant C ' instead?
2
(2) Why and how is x 2 changed into u + 1 ?
Example 4
Suppose x > 0 .
Find (a) ∫ e
ax + b
dx (b) ∫ (2ax + b)e
ax 2 + bx + c
2
1
dx (c) ∫ 2 e x dx (d)
x
∫
ln x
dx .
x
Solution:
1 ax +b
e ax +b
e
d
(
ax
+
b
)
=
+C
a∫
a
(a)
ax + b
∫ e dx =
(b)
ax
∫ (2ax + b)e
(c)
1
1
1
2
∫ x 2 e x dx = − 2 ∫ e x d  x  = − 2 e x + C
(d)
ln x
(ln x ) 2
dx
=
ln
xd
(ln
x
)
=
+C
∫ x
∫
2
2
2
+ bx + c
dx = ∫ e ax
2
2
+ bx + c
d (ax 2 + bx + c) = e ax
2
5
2
+ bx + c
+C
Example 5
Find ∫ x 2 ( x + 1) 2013 dx .
Solution:
Let u = x + 1 , then du = dx .
∫ x ( x + 1) dx
= ∫ (u − 1) u du
= ∫ (u − 2u + 1)u du
= ∫ (u
− 2u
+u
2
2013
2
2013
2
2013
2015
2014
2013
) du
u 2016 2u 2015 u 2014
−
+
+C
2016 2015 2014
( x + 1) 2016 2( x + 1) 2015 ( x + 1) 2014
=
−
+
+C
2016
2015
2014
=
Example 6
x2
x +1
Find (a) ∫
dx .
dx (b) ∫
x −1
x −1
Solution:
(a)
x +1
∫ x − 1 dx = ∫
x −1+ 2
dx
x −1
2 
 x −1
= ∫
+
 dx
 x −1 x −1 
1
= ∫ dx + 2∫
d ( x − 1)
x −1
= x + 2 ln x − 1 + C
We can write
∫ 1dx
simply as
∫ dx , omitting the constant 1.
(b) Let u = x − 1 , then du = dx .
x2
∫ x − 1 dx
(u + 1) 2
=∫
du
u
u 2 + 2u + 1
du
=∫
u
1
= ∫ (u + 2 + )du
u
2
u
=
+ 2u + ln u + C
2
( x − 1) 2
=
+ 2( x − 1) + ln x − 1 + C
2
( x − 1) 2
=
+ 2 x + ln x − 1 + C '
2
6
Example 7 (Long Division & Partial Fractions)
x 6 − x 5 − x 4 − x 3 − 6x 2 + 8x + 1
g ( x)
, where f (x ) and g (x ) are
≡ f ( x) + 2
2
x − 3x + 2
x − 3x + 2
two polynomials with deg f ( x) ≤ 4 and deg g ( x ) < 2 . Find f (x ) and g (x ) .
(a) Let
(b) Let
g ( x)
A
B
, where A and B are constants. Find A and B.
≡
+
x − 3x + 2 x − 1 x − 2
2
(c) Hence, find
x 6 − x 5 − x 4 − x 3 − 6x 2 + 8x + 1
dx .
∫
x 2 − 3x + 2
Solution:
Points to note:
1. The notation ‘deg’ is used to refer to the
degree of a polynomial. For instance, let
(a) Here, we perform long division.
h( x ) = x 5 − 2 x 2 + 1 , then deg h( x ) is
1 + 2 +3+ 4
1 − 3 + 2 1 −1 −1 −1 − 6 + 8 +1
equal to 5.
2. This type of long division without
1 −3+ 2
variables written is known as the method of
+ 2 − 3 −1 − 6 + 8 +1
detached coefficients.
+2−6+4
3. The method employed in (b), i.e. to break
+ 3 − 5 − 6 + 8 +1
an algebraic fraction into sum of fractions
+3−9 +6
with denominators of smaller degrees, is
+ 4 − 12 + 8 + 1
called resolving an algebraic fraction into
+ 4 − 12 + 8
partial fractions.
+1
x 6 − x 5 − x 4 − x 3 − 6x 2 + 8x + 1
1
= x 4 + 2 x 3 + 3x 2 + 4 x + 2
.
2
x − 3x + 2
x − 3x + 2
Thus, f ( x ) ≡ x 4 + 2 x 3 + 3 x 2 + 4 x and g ( x ) ≡ 1 .
Then, we have
(b) From (a), g ( x ) ≡ 1 .
1
A
B
≡
+
x − 3x + 2 x − 1 x − 2
1 ≡ A( x − 2) + B( x − 1)
2
Put x = 1 , we have 1 = A(1 − 2) + B (1 − 1) , i.e. A = −1 .
Put x = 2 , we have 1 = A(2 − 2) + B ( 2 − 1) , i.e. B = 1 .
(c) From above, we have
x 6 − x 5 − x 4 − x 3 − 6x 2 + 8x + 1
1
1
= x 4 + 2 x 3 + 3x 2 + 4 x +
−
2
x − 2 x −1
x − 3x + 2
1
1 
x 6 − x 5 − x 4 − x 3 − 6 x 2 + 8x + 1

dx = ∫  x 4 + 2 x 3 + 3 x 2 + 4 x +
−
dx
2
∫
x − 2 x −1
x − 3x + 2

x5 x4
=
+
+ x 3 + 2 x 2 + ln x − 2 − ln x − 1 + C
5
2
5
x
x4
x−2
=
+
+ x 3 + 2 x 2 + ln
+C
5
2
x −1
7
Example 8 (Reduction Formula)
Let y = x(ln x ) n , where x > 0 .
(a) Find
dy
.
dx
(b) Let I n = ∫ (ln x) n dx , where n ≥ 0 .
Using the result in (a), or otherwise, show that I n = x(ln x) n − nI n −1
(c) Hence, find I 4 .
Solution:
(a)
dy
1
= x ⋅ n(ln x ) n −1 ⋅ + (ln x ) n ⋅ (1) = n(ln x ) n −1 + (ln x ) n
dx
x
(b) Integrating both sides w.r.t. x, we have
y = ∫ n(ln x ) n−1 dx + ∫ (ln x ) n dx
x (ln x ) n = n ∫ (ln x ) n−1 dx + ∫ (ln x ) n dx
∫ (ln x)
n
dx = x(ln x ) n − n ∫ (ln x ) n −1 dx
I n = x(ln x ) n − nI n −1
(c) Using the result in (b) repeatedly, we have
I 4 = x (ln x ) 4 − 4 I 3
[
= x (ln x ) 4 − 4 x (ln x ) 3 − 3I 2
]
[
= x (ln x ) 4 − 4 x (ln x ) 3 + 12 x (ln x ) 2 − 2 I 1
]
= x (ln x ) 4 − 4 x (ln x ) 3 + 12 x (ln x ) 2 − 24[x ln x − I 0 ]
= x (ln x ) 4 − 4 x (ln x ) 3 + 12 x (ln x ) 2 − 24 x ln x + 24 ∫ (ln x ) 0 dx
= x (ln x ) 4 − 4 x (ln x ) 3 + 12 x (ln x ) 2 − 24 x ln x + 24 x + C
Notes:
(1) The method employed in this question is to establish a reduction formula, which
reduces an integral from a higher to lower power, with the same form. A reduction
formula can be used repeatedly to lower the power of the integral, until it can be
integrated easily. In this way, the integrals in certain forms, no matter how high the
power is, can be found.
(2) We can find out a reduction formula by differentiation. However, we usually use a
technique called integration by parts to find them out. This will be introduced in
Section 3.5. (Refer to Example 34 for details)
8
3.3. Integration of Trigonometric Functions
Trigonometric functions can be differentiated. Similarly, they can also be integrated.
3.3.1.
Basic Integration of Trigonometric Functions
From differentiation, we have the following basic results.
(1)
∫ sin xdx = − cos x + C
(2)
∫ cos xdx = sin x + C
(3)
∫ sec
(4)
∫ csc
(5)
∫ sec x tan xdx = sec x + C
(6)
∫ csc x cot xdx = − csc x + C
2
xdx = tan x + C
2
xdx = − cot x + C
For other integrals of trigonometric functions, different trigonometric identities are useful.
Example 9
Find (a) ∫ sin 2
x
dx (b) ∫ tan 2 xdx .
2
Solution:
(a)
∫ sin
(b)
∫ tan
2
x
1 − cos x
1
1
1
1
1
1
dx = ∫
dx = ∫ dx − ∫ cos xdx = x − ∫ cos xdx = x − sin x + C
2
2
2
2
2
2
2
2
2
xdx = ∫ (sec 2 x − 1) dx = tan x − x + C
(Can you find
3.3.2.
∫ cos
2
x
dx and
2
∫ cot
2
xdx ? Give them a try!)
Integration of Trigonometric Functions with Method of Substitution
The method of substitution can also be used in finding integrals of trigonometric functions.
Example 10
Point to note:
Find ∫ sin 3x cos xdx .
Generally, we have the following formulas
for constants a and b, with a ≠ 0 :
Solution:
1
∫ sin(ax + b)dx = − a cos(ax + b) + C
∫ sin 3x cos xdx
1
1
[sin(3x + x) + sin(3x − x)]dx
2∫
1
= ∫ (sin 4 x + sin 2 x)dx
2
1 1
1 1
= ⋅ ∫ sin 4 xd (4 x) + ⋅ ∫ sin 2 xd (2 x)
2 4
2 2
1
1
= − cos 4 x − cos 2 x + C
8
4
∫ cos(ax + b)dx = a sin(ax + b) + C
=
They can be proven easily by differentiation
or method of substitution, with substitution
u = ax + b . They are widely accepted
and are not required to be proven again
in your calculations.
9
Example 11
(a) Prove that tan x =
1 − cos 2 x
.
sin 2 x
(b) Hence, or otherwise, find
∫
1 − cos 2 x
dx .
sin 2 x
Solution:
(a)
RHS =
1 − cos 2 x
2
1 − cos 2 x
sin 2 x
sin x
=
⋅
=
=
= tan x = LHS
sin 2 x
2 sin x cos x
2
sin x cos x cos x
Thus, we have tan x =
(b)
∫
1 − cos 2 x
.
sin 2 x
1 − cos 2 x
sin x
1
dx = ∫ tan xdx = ∫
dx = − ∫
d (cos x ) = − ln cos x + C (or ln sec x + C )
sin 2 x
cos x
cos x
Example 12
Find ∫ sec xdx .
Solution:
∫ sec xdx
=∫
sec x (sec x + tan x )
dx
sec x + tan x
=∫
d (sec x + tan x )
sec x + tan x
(Here, u = sec x + tan x and du = sec x (sec x + tan x ) dx .)
= ln sec x + tan x + C
Notes:
(1) Students are strongly recommended to recite the substitution used in Example 12, as well
as the way to find this integral.
(2) Alternatively, you can find the integral using partial fractions as follows:
1
cos x
1
1
dx = ∫
d (sin x) = ∫
du (for u = sin x )
2
2
x
1 − sin x
1− u2
∫ sec xdx = ∫ cos x dx = ∫ cos
We can let
1
A
B
. Solving the simultaneous equations set up by
≡
+
2
1+ u 1− u
1− u
comparing like terms, we can get A =
Then, ∫ sec xdx =
1
1
and B = . (Refer to Example 7 for details)
2
2
1
1
1
1
1
1
1 1 + sin x
du + ∫
du = ln 1 + u − ln 1 − u + C = ln
+C .
∫
2 1+ u
2 1− u
2
2
2 1 − sin x
(3) Can you find ∫ csc xdx ? Give it a try! (Hint: Let u = csc x + cot x )
10
Example 13 (Subsidiary Angle)
(a) Let 0 < θ <
π
. By differentiating ln(sec θ + tan θ ) , show that ∫ sec θdθ = ln(secθ + tan θ ) + C ,
2
where C is a constant.
(b) Let cos θ − sin θ = r cos(θ + α ) , where r and α are constants with r > 0 and −
π
π
<α < .
2
2
Find r and α .
(c) Hence, find
1
∫ cos θ − sin θ dθ .
Solution:
(a)
d
1
ln(secθ + tan θ ) =
(sec θ tan θ + sec 2 θ ) = sec θ
dθ
sec θ + tan θ
Integrating both sides w.r.t. θ , we have
∫ secθdθ = ln(secθ + tan θ ) + C
(b) cos θ − sin θ = r cos(θ + α ) = r cos θ sin α − r cos α sin θ
Comparing like terms, we have
......................(1)
r sin α = 1

− r cos α = −1.......................( 2)
(1)
: tan α = 1
( 2)
We get α =
π
.
4
(1) 2 + ( 2) 2 : r 2 (sin 2 α + cos 2 α ) = 12 + 12
We get r = 2 .
π

(c) From (b), we have cos θ − sin θ = 2 cosθ +  .
4

1
1
1
∫ cos θ − sin θ dθ = 2 ∫  π  dθ
cosθ + 
4

1
π 
π

=
sec θ + d θ + 
∫
4 
4
2

=
2  
π
π 

ln sec θ +  + tan  θ +  + C
2  
4
4 

Notes:
(1) The form r sin( x ± α ) or r cos( x ± α ) is called the subsidiary angle form. It expresses the sum
(or difference) of two sine (or cosine) functions (with any non-zero coefficients) as a product
of a non-zero real number and a sine (or cosine) function.
(2) Subsidiary angle form is NOT required in HKDSE M2 Examination. However, question
format as in Example 13 is possible to appear in the examination.
11
Example 14
sin(ln x)
dx (b) ∫ sin x sec 2 (cos x )dx (c)
x
Find (a) ∫
∫ x csc( x
2
+ 1) cot( x 2 + 1) dx .
Solution:
(a)
∫
sin(ln x)
dx
x
= ∫ sin(ln x )d (ln x )
= − cos(ln x ) + C
(b)
∫ sin x sec
2
(cos x )dx
= − ∫ sec 2 (cos x )d (cos x )
= − tan(cos x ) + C
(c)
∫ x csc( x
2
+ 1) cot( x 2 + 1) dx
1
csc( x 2 + 1) cot( x 2 + 1)d ( x 2 + 1)
∫
2
1
= − csc( x 2 + 1) + C
2
=
3.3.3.
Integration of Special Types of Trigonometric Functions
The integrals of some trigonometric functions can be found easily by certain methods.
(Here, p and q are positive integers.)
Type A:
∫ sin
p
x cos q xdx
Example 15
Find (a) ∫ sin 5 x cos xdx (b) ∫ sin 6 x cos 3 xdx (c) ∫ sin 6 x cos 4 xdx .
Solution:
(a)
5
5
∫ sin x cos xdx = ∫ sin xd (sin x) =
(b)
∫ sin
6
sin 6 x
+C
6
x cos 3 xdx
= ∫ sin 6 x ⋅ cos 2 x ⋅ cos xdx
= ∫ sin 6 x(1 − sin 2 x )d (sin x )
= ∫ sin 6 xd (sin x ) − ∫ sin 8 xd (sin x )
=
sin 7 x sin 9 x
−
+C
7
9
12
(c) We first express sin 6 x cos 4 x as a sum of cos nx , where n is an integer.
sin 6 x cos 4 x = (sin x cos x) 4 sin 2 x
4
1
  1 − cos 2 x 
=  sin 2 x  

2
2
 

1
= sin 4 2 x(1 − cos 2 x)
32
2
1  1 − cos 4 x 

 (1 − cos 2 x)
32 
2

1
=
(1 − 2 cos 4 x + cos 2 4 x)(1 − cos 2 x)
128
1 
1 + cos 8 x 
=
1 − 2 cos 4 x +
(1 − cos 2 x)
128 
2

1
=
(3 − 4 cos 4 x + cos 8 x )(1 − cos 2 x)
256
1
=
(3 − 3 cos 2 x − 4 cos 4 x + 4 cos 2 x cos 4 x + cos 8 x − cos 2 x cos 8 x)
256
1 
1
1

=
 3 − 3 cos 2 x − 4 cos 4 x + 2 cos 2 x + 2 cos 6 x + cos 8 x − cos10 x − cos 6 x 
256 
2
2

1
=
(6 − 2 cos 2 x − 8 cos 4 x + 3 cos 6 x + 2 cos 8 x − cos10 x)
512
=
∫ sin
6
1
(6 − 2 cos 2 x − 8 cos 4 x + 3 cos 6 x + 2 cos 8 x − cos10 x )dx
512 ∫
1 
1
1
1

=
 6 x − sin 2 x − 2 sin 4 x + sin 6 x + sin 8 x − sin 10 x  + C
512 
2
4
10

3
1
1
1
1
1
=
x−
sin 2 x −
sin 4 x +
sin 6 x +
sin 8 x −
sin 10 x + C
256
512
256
1024
2048
5120
x cos 4 xdx =
Notes:
(1) If at least one of p or q is odd, we use the substitution u = sin x or u = cos x , in order to
convert the integral into a form only with the substitution used. In Examples 13(a) and 13(b),
we used u = sin x as the substitution and the whole integral was expressed in terms of sin x .
(2) If both p and q are even, we use the identities sin 2 x =
1 − cos 2 x
1 + cos 2 x
and cos 2 x =
2
2
repeatedly to lower the power of the integral. The integral is eventually expressed as a sum of
sine and cosine functions, and can then be integrated easily.
13
Type B:
∫ tan
p
x sec q xdx
Example 16
Find (a) ∫ tan 5 x sec 2 xdx (b) ∫ tan 2 x sec 4 xdx (c) ∫ tan 3 x sec 5 xdx .
Solution:
(a)
∫ tan
5
x sec 2 xdx
= ∫ tan 5 xd (tan x)
=
(b)
tan 6 x
+C
6
∫ tan
2
x sec 4 xdx
= ∫ tan 2 x sec 2 x ⋅ sec 2 xdx
= ∫ tan 2 x (tan 2 x + 1) d (tan x )
= ∫ tan 4 xd (tan x ) + ∫ tan 2 xd (tan x )
=
(c)
tan 5 x tan 3 x
+
+C
5
3
∫ tan
3
x sec 5 xdx
= ∫ tan 2 x sec 4 x ⋅ sec x tan xdx
= ∫ (sec 2 x − 1) sec 4 xd (sec x )
= ∫ sec 6 xd (sec x ) − ∫ sec 4 xd (sec x )
=
sec 7 x sec 5 x
−
+C
7
5
Notes:
(1) If q is even, we use the substitution u = tan x and convert the whole integral in terms of tan x .
Besides this, sec n x (where n is an even number) can be expressed in terms of tan x using the
relation sec 2 x = 1 + tan 2 x .
(2) If both p and q are odd, we use the substitution u = sec x and convert the whole integral in
terms of sec x . Besides this, tan n x (where n is an even number) can be expressed in terms of
sec x using the relation tan 2 x = sec 2 x − 1 .
(3) It is much more difficult to find the integral if p is even and q is odd. A new technique called
‘integration by parts’ is required, and it will be introduced in Section 3.5.
(Refer to Example 33 on p. 33 for details.)
(4) Integrals in the form
∫ cot
p
x csc q xdx are found in similar way.
14
Type C:
∫ sin
xdx ,
n
∫ cos
n
xdx , etc.
The techniques employed are similar to those used in finding Type A and Type B integrals.
Example 17
(a) Find (i) ∫ sin 4 xdx (ii) ∫ cos 5 xdx (iii) ∫ tan 4 xdx (iv) ∫ cot 5 xdx .
n −1
1
(b) By differentiating sin n −1 x cos x , show that ∫ sin n xdx = − sin n −1 x cos x +
sin n − 2 xdx ,
n
n ∫
where n ≥ 2 . Hence, find
∫ sin
4
xdx .
Solution:
∫ sin
(a) (i)
4
xdx
2
 1 − cos 2 x 
= ∫
 dx
2


1
= ∫ (1 − 2 cos 2 x + cos 2 2 x)dx
4
1 
1 + cos 4 x 
= ∫ 1 − 2 cos 2 x +
 dx
4 
2

1
= ∫ (3 − 4 cos 2 x + cos 4 x)dx
8
3
1
1
= x − sin 2 x + sin 4 x + C1
8
4
32
(ii)
∫ cos
5
We try to express the integral
as a sum of cosine functions.
xdx
We try to express the integral
in terms of sin x .
= ∫ (1 − sin 2 x) 2 cos xdx
= ∫ (1 − 2 sin 2 x + sin 4 x)d (sin x)
2
1
= sin x − sin 3 x + sin 5 x + C
3
5
(iii)
∫ tan
4
xdx
= ∫ (sec 2 x − 1) tan 2 xdx
= ∫ sec 2 x tan 2 xdx − ∫ tan 2 xdx
= ∫ tan 2 xd (tan x) − ∫ (sec 2 x − 1) dx
=
tan 3 x
− tan x + x + C
3
15
∫ cot
(iv)
5
xdx
= ∫ (csc 2 x − 1) 2 ⋅ cot xdx
= ∫ csc 4 x cot xdx − 2 ∫ csc 2 x cot xdx + ∫ cot xdx
= − ∫ csc 3 xd (csc x ) + 2 ∫ csc xd (csc x) + ∫
cos x
dx
sin x
csc 4 x
1
+ csc 2 x + ∫
d (sin x )
4
sin x
csc 4 x
=−
+ csc 2 x + ln sin x + C
4
=−
(b) Let y = sin n −1 x cos x .
dy
= sin n −1 x ( − sin x ) + (cos x )(n − 1)(sin n− 2 x )(cos x)
dx
= − sin n x + ( n − 1) sin n −2 x cos 2 x
= − sin n x + ( n − 1) sin n −2 x(1 − sin 2 x )
= ( n − 1) sin n −2 x − n sin n x
Integrating both sides w.r.t. x, we have
y = (n − 1) ∫ sin n − 2 xdx − n ∫ sin n xdx
n ∫ sin n xdx = (n − 1) ∫ sin n − 2 xdx − sin n −1 cos x
∫ sin
n
∫ sin
4
1
n −1
xdx = − sin n −1 x cos x +
sin n − 2 xdx
∫
n
n
1
3
xdx = − sin 3 x cos x + ∫ sin 2 xdx
4
4
1
3 1
1

= − sin 3 x cos x +  − sin x cos x + ∫ (sin x ) 0 dx 
4
4 2
2

1
3
3
= − sin 3 x cos x − sin x cos x + x + C 2
4
8
8
An integral may take up different forms. We obtain different results for the same integral ∫ sin 4 xdx .
If we subtract the result in (a)(i) from that in (b), we get:
3
3  3
1
1
 1 3

 − sin x cos x − sin x cos x + x  −  x − sin 2 x + sin 4 x 
8
8  8
4
32
 4

11
1
1
 1 − cos 2 x  3
= −  sin 2 x 
 − sin 2 x + sin 2 x − sin 2 x cos 2 x
42
2
4
16

 16
1
1
3
1
1
= − sin 2 x + sin 2 x cos 2 x − sin 2 x + sin 2 x − sin 2 x cos 2 x
16
16
16
4
16
=0
It follows that C1 = C2. There is no difference between the two integrals, and they are solely equivalent.
16
3.4. Integration by Trigonometric Substitution
Integration of trigonometric functions can be helpful in finding certain indefinite integrals. Before
learning how to do so, we have to learn about inverse trigonometric functions.
3.4.1.
Inverse Trigonometric Functions
We have come across with a lot of trigonometric functions, say, y = sin x . To express x in terms of y,
we define x = sin −1 y (or x = arcsin y ), where sin −1 y is read as ‘arcsine y’.
(a) Principal Values
2
π 3π 9π
π
, we may have x = , ,
,... (in fact, x = nπ + ( −1) n ,
2
4 4 4
4
where n is an integer). A function should give one and only one output for every input (e.g.
for x = sin −1 y , if x is a function of y, then for any value of y, there is one and only one
Consider y = sin x . For y =
corresponding value of x). To make sure the relationship, we define an interval called
‘principal values’ to restrict the value of x.
Refer to the following illustration:
Let y = sin −1 x . (Note that x and y is exchanged here, unlike the above example)
π
π
 π π
For arcsine functions, the principal values are − ,  (i.e. − ≤ x ≤ ).
2
2
 2 2
When x =
1
π
1 π
, we have y = , i.e. sin −1 = .
2
6
2 6
When x = −1 , we have y = −
π
π
, i.e. sin −1 ( −1) = − .
2
2
Note: Our calculators are user-friendly. When you enter inverse trigonometric functions, it will
automatically show the principal value as the answer.
The principal values of different functions are shown in the table below:
Function
sin −1 x
−1
cos x
tan −1 x
csc −1 x
sec −1 x
cot
Note: cot
−1
x
Principal Values
Point to note:
 π π
− 2 , 2 


[0, π ]
[ ] refers to closed interval (i.e. including the
 π π
− , 
 2 2
Examples:
limits) while ( ) refers to open interval (i.e.
excluding the limits).
π

− π ,− 2  or


 π
0, 2 


π

− π ,− 2  or


(0, π )
 π
0, 2 


π
= 0.
2
17
π
π
 π π
(1) x ∈ − ,  means − ≤ x ≤ .
2
2
 2 2
π
π
 π π
(2) x ∈  − ,  means − < x < .
2
2
 2 2
 π π
π
π
(3) x ∈  − ,  means − < x ≤ .
2
2
 2 2
Example 18
Find (a) sin −1 0.75 (b) cos −1
π
(c) cot −1 3.3 , in radian measures, correct to 3 decimal places.
4
Solution:
(a)
sin −1 0.75 ≈ 0.848
(i.e. sin 0.848 ≈ 0.75 )
(b) cos −1
Let x = cot −1 3.3 .
Then, we have
cot x = 3.3
π
≈ 0.667
4
1
= 3. 3
tan x
1
tan x =
3. 3
1
x = tan −1
3 .3
π
(i.e. cos 0.667 ≈ )
4
(c) We cannot enter ‘ cot −1 ’ directly into our calculators.
Therefore, we use the following method.
 1 
cot −1 3.3 = tan −1 

 3.3 
≈ 0.294
(i.e. cot 0.294 ≈ 3.3 )
 1 
i.e. cot −1 3.3 = tan −1 

 3 .3 
(b) Graphs of Inverse Trigonometric Functions
The following shows the graphs of the six trigonometric functions and those of their inverse
functions for comparison:
y
y
y=sin x
y=sin−1 x
x
O
O
−1
y
1
x
y
π
y=cos x
O
π
2
x
−1
18
O
y=cos−1 x
1
x
y
y
y=tan x
y=tan−1 x
x
O
x
O
y
y
π
2
y=csc x
x
O
y=csc−1 x
1
−1
O
x
π
−
2
y
y=sec−1 x
y
x
y=sec x
1
−1
x
O
−π
O
y
y
y=cot x
O
x
π
2
y=cot−1 x
O
Figure 2 The graphs of the six trigonometric functions and their inverse functions
19
x
3.4.2.
Basic Integration by Trigonometric Substitution
Trigonometric substitutions are useful in finding integrals involving a 2 − x 2 , a 2 + x 2 and x 2 − a 2 ,
where a is a constant. The following examples will illustrate how we use the substitution to find
these integrals.
Type A: Integrals involving a 2 − x 2
We use the substitution x = a sin θ , where −
π
π
<θ < .
2
2
Example 19
Find
∫
1
25 − x 2
dx .
Solution:
Let x = 5 sin θ , then dx = 5 cosθdθ .
1
1
∫ 25 − x 2 dx = ∫ 25 − (5 sin θ ) 2 ⋅ 5 cos θdθ
5 cos θ
=∫
dθ
5 cos 2 θ
cos θ
=∫
dθ
cos θ
= ∫ dθ
=θ +C
= sin −1
x
+C
5
Notes:
(1) In the range −
π
π
< θ < , cosθ is positive. As a result,
2
2
cos 2 θ = cosθ .
(Normally, we do not have to specify the range of θ in our calculations.)
(2) We can also use the substitution x = 5 cosθ as follows:
Let x = 5 cosθ , then dx = −5 sin θdθ .
1
1
∫ 25 − x 2 dx = ∫ 25 − (5 cos θ ) 2 ⋅ (−5 sin θ )dθ
= −∫
5 sin θ
dθ
5 sin 2 θ
sin θ
= −∫
dθ
sin θ
= − ∫ dθ
= −θ + C
= − cos −1
x
+C
5
20
Example 20
Find ∫ 9 − x 2 dx .
Solution:
Let x = 3 sin θ , then dx = 3 cosθdθ .
∫
9 − x 2 dx = ∫ 9 − (3 sin θ ) 2 ⋅ 3 cos θdθ
= 9 ∫ 1 − sin 2 θ cos θdθ
= 9 ∫ cos 2 θdθ
9
(1 + cos 2θ ) dθ
2∫
9
9
= θ + sin 2θ + C
2
4
=
Now, we need to express the result in terms of x.
θ = sin −1
x
3
3
sin 2θ = 2 sin θ cosθ
[We construct a right-angled triangle as shown on the right such that sin θ =
[Length of adjacent side =
2
 x  9 − x
sin 2θ = 2 
3
 3 
∫
9 − x 2 dx =
=
θ
32 − x 2 = 9 − x 2 ]
9 − x2
2
[Then, we have cosθ =
x
.]
3
9− x
.]
3
 2x 9 − x 2
=

9

9 −1 x 9  2 x 9 − x 2
sin
+
2
3 4 
9

+C


9 −1 x 1
sin
+ x 9 − x2 + C
2
3 2
Notes:
(1) Sometimes, we may need to further perform integration of trigonometric functions after using
a trigonometric substitution.
(2) The steps of finding cosθ (enclosed by [ ]) can be skipped, as long as the right-angled triangle
(with the lengths of three sides clearly indicated) is clearly drawn.
21
x
Type B: Integrals involving a 2 + x 2
We use the substitution x = a tan θ , where −
π
π
<θ < .
2
2
Example 21
It is given that
Find (a) ∫
∫ sec xdx = ln sec x + tan x + C . (Refer to Example 12 for details)
1
1
dx .
dx (b) ∫
x +3
x2 + 3
2
Solution:
(a) Let x = 3 tan θ , then dx = 3 sec 2 θdθ .
1
3 sec 2 θ
dx
=
∫ x 2 + 3 ∫ ( 3 tan θ ) 2 + 3 dθ
=∫
3 sec 2 θ
dθ
3(tan 2 θ + 1)
=
3 sec 2 θ
dθ
3 ∫ sec 2 θ
3
θ + C1
3
3
x
tan −1
=
+ C1
3
3
=
(b) Let x = 3 tan θ , then dx = 3 sec 2 θdθ .
∫
1
x2 + 3
dx = ∫
=∫
3 sec 2 θ
dθ
( 3 tan θ ) 2 + 3
3 sec 2 θ
dθ
3 ⋅ tan 2 θ + 1
sec 2 θ
=∫
dθ
sec θ
x2 + 3
= ∫ sec θdθ
3
= ln sec θ + tan θ + C '
= ln
x2 + 3
3
x
+
3
θ
+ C'
x
= ln x 2 + 3 + x − ln 3 + C '
= ln x 2 + 3 + x + C ' ' , where C ' ' = C '− ln 3
Note:
As −
π
π
<θ < ,
2
2
sec 2 θ = secθ , but not − secθ .
22
Example 22
Find
∫
x3
x2 + 4
dx by (a) a trigonometric substitution (b) an algebraic substitution.
Solution:
(a) Let x = 2 tan θ , then dx = 2 sec 2 θdθ .
x3
( 2 tan θ ) 3
2
dx
=
∫ x2 + 4
∫ (2 tan θ ) 2 + 4 ⋅ 2 sec θdθ
=∫
8 tan 3 θ
⋅ 2 sec 2 θdθ
2
x2 + 4
2 sec θ
x
= 8∫ tan 3 θ sec θdθ
= 8∫ (sec 2 θ − 1) d (sec θ )
θ
8
= sec 3 θ − 8 sec θ + C
3
3
 x2 + 4 
8  x 2 + 4 
+C
= 
− 8



3 
2
2



2
3
1
= ( x 2 + 4) 2 − 4 x 2 + 4 + C
3
1 2
=
x + 4 ( x 2 − 8) + C
3
(b) Let u = x 2 + 4 , then du = 2 xdx .
∫
x3
x2 + 4
x3
1
du
u 2x
1 u−4
= ∫ 1 ⋅ du
2
u2
1
1
−
1
= ∫ (u 2 − 4u 2 )du
2
3
1
1 2
= u − 4u 2 + C
3
3
1
1
= ( x 2 + 4) 2 − 4( x 2 + 4) 2 + C
3
1 2
=
x + 4 ( x 2 − 8) + C
3
dx = ∫
⋅
Notes:
An integral can be found by different methods. In this example, we used trigonometric substitution
and algebraic substitution, and we still arrived at the same result.
23
Type C: Integrals involving x 2 − a 2
We use the substitution x = a secθ , where 0 < θ <
π
π
or − π < θ < − .
2
2
Example 23
It is given that
Find (a) ∫
∫ sec xdx = ln sec x + tan x + C . (Refer to Example 12 for details)
1
x x2 − 9
1
dx (b) ∫
x2 − 9
dx .
Solution:
(a) Let x = 3 secθ , then dx = 3 secθ tan θdθ .
∫x
1
2
x −9
dx = ∫
=∫
3 sec θ tan θ
dθ
3 sec θ (3 sec θ ) 2 − 9
tan θ
dθ
3 tan 2 θ
1 tan θ
= ∫
dθ
3 tan θ
1
= θ +C
3
1
x
= sec −1 + C1
3
3
(b) Let x = 3 secθ , then dx = 3 secθ tan θdθ .
∫
1
2
x −9
dx = ∫
=∫
3 sec θ tan θ
dθ
(3 sec θ ) 2 − 9
sec θ tan θ
dθ
tan 2 θ
x
x2 − 9
= ∫ sec θdθ
= ln sec θ + tan θ + C '
x
= ln +
3
x2 − 9
+ C'
3
θ
= ln x + x 2 − 9 + C ' ' , where C ' ' = C '− ln 3
Notes:
In the range 0 < θ <
π
π
or − π < θ < − ,
2
2
tan 2 θ = tan θ , but not − tan θ .
24
3
3.4.3.
Variations of Integration by Trigonometric Substitution
Example 24
Find
cos x
∫ 1 + sin
2
x
dx .
Solution:
Let sin x = tan θ , then cos xdx = sec 2 θdθ .
cos x
sec 2 θ
∫ 1 + sin 2 x dx = ∫ 1 + tan 2 θ dθ
= ∫ dθ
=θ +C
= tan −1 (sin x ) + C
Example 25 (Method of Completing the Square)
Find ∫
1
dx .
2
− x + 12 x − 20
Solution:
We first express − x 2 + 12 x − 20 in the form of a 2 − ( x − b) 2 , where a and b are constants.
− x 2 + 12 x − 20 = −( x 2 − 12 x) − 20
2
2
 2
 12  
 12 
= −  x − 12 x +    − 20 +  
 2
 2  

= 16 − ( x − 6) 2
= 4 2 − ( x − 6) 2
Let x − 6 = 4 sin θ , then dx = 4 cosθdθ .
∫
1
− x 2 + 12 x − 20
dx = ∫
=∫
1
dx
4 2 − ( x − 6) 2
4 cos θ
dθ
2
4 − (4 sin θ ) 2
cos θ
dθ
cos θ
=θ +C
x−6
= sin −1
+C
4
=∫
Notes:
The method of completing the square is used in converting quadratic expressions into the form
required for using trigonometric substitutions. We usually use this method when we encounter a
quadratic denominator, while the numerator is a constant.
25
3.5. Integration by Parts
3.5.1. Introduction
There are still some integrals that cannot be found using the above techniques, e.g. ∫ x sin xdx ,
∫ x ln xdx
and
∫e
x
cos xdx , etc. which are mostly in the form of the products of two functions.
Here, we need a new technique called integration by parts to deal with these integrals.
Let u = f (x ) and v = g (x ) be two differentiable functions.
Then, we have
∫ udv = uv − ∫ vdu .
The proof is amazingly simple with the help of the product rule of differentiation.
Proof:
d
(uv) = uv '+ vu '
dx
Integrating both sides w.r.t. x, we have
uv = ∫ uv ' dx + ∫ vu ' dx
= ∫ udv + ∫ vdu
i.e. ∫ udv = uv − ∫ vdu .
The following example illustrates how we can use integration by parts to find an integral.
Example 26
Find ∫ ln xdx .
Solution:
Consider ∫ ln xdx .
Take u = ln x and v = x .
We have du =
1
dx and dv = dx .
x
∫ ln xdx = ∫ udv
= uv − ∫ vdu
In fact, it can be written in a simpler way:
∫ ln xdx = x ln x − ∫ xd (ln x)
1
= x ln x − ∫ x ⋅ dx
x
1
= x ln x − ∫ x ⋅ dx
x
= x ln x − ∫ dx
= x ln x − ∫ dx
= x ln x − x + C
= x ln x − x + C
26
3.5.2.
Common Integrals involving Integration by Parts
There are many integrals that can be found using integration by parts. Some of them are listed here.
Type A:
∫x
n
sin xdx or
∫x
n
cos xdx
Example 27
Find (a) ∫ x sin xdx (b) ∫ x cos xdx (c) ∫ x 2 sin 2
x
dx .
2
Solution:
∫ x sin xdx
(a)
Here, u = x , v = cos x ,
du = dx and dv = − sin xdx .
= − ∫ xd (cos x )
= − x cos x + ∫ cos xdx
= − x cos x + sin x + C
∫ x cos xdx
(b)
Here, u = x , v = sin x ,
du = dx and dv = cos xdx .
= ∫ xd (sin x )
= x sin x − ∫ sin xdx
= x sin x + cos x + C
∫x
(c)
=
=
=
=
=
=
2
sin 2
x
dx
2
1 2
x (1 − cos x )dx
2∫
1 2
1
x dx − ∫ x 2 cos xdx
∫
2
2
3
x
1
− ∫ x 2 d (sin x )
6 2
x3 1 2
1
− x sin x + ∫ sin xd ( x 2 )
6 2
2
3
x
1
− x 2 sin x + ∫ x sin xdx
6 2
x3 1 2
− x sin x − x cos x + sin x + C
6 2
Use integration by parts
repeatedly here (as in part
(a)) to find
∫ x sin xdx .
Notes:
Integration by parts can be used repeatedly to lower the power of some integrals. In DSE M2
examination, however, you are required to use this technique at most twice only to find an integral.
27
∫x
Type B:
n
ln xdx
Example 28
Find ∫ x n ln xdx for (a) n ≠ −1 (b) n = −1 .
Solution:
∫x
(a)
n
ln xdx
1
ln xd ( x n+1 )
∫
n +1
1 n +1
1
=
x ln x −
x n +1 d (ln x)
∫
n +1
n +1
1 n +1
1
=
x ln x −
x n dx
∫
n +1
n +1
1 n +1
1
=
x ln x −
x n +1 + C
2
n +1
(n + 1)
=
∫x
(b)
=∫
−1
ln xdx
ln x
dx
x
= ∫ ln xd (ln x)
(ln x) 2
+C
2
*Think About*:
=
Can we find the integral by expressing
Type C:
∫x
n
∫x
n
ln xdx as x n +1 ln x − ∫ xd ( x n ln x ) ?
e x dx
Example 29
Find (a) ∫ xe x dx (b) ∫ x 2 e 2 x dx .
Solution:
(a)
∫ xe
x
dx
= ∫ xd (e x )
= xe x − ∫ e x dx
= xe x − e x + C
28
∫x
(b)
=
=
=
=
=
=
2
e 2 x dx
1 2
x d (e 2 x )
2∫
1 2 2x 1 2x
x e − ∫ e d (x 2 )
2
2
1 2 2x
x e − ∫ xe 2 x dx
2
1 2 2x 1
x e − ∫ xd (e 2 x )
2
2
1 2 2x 1 2x 1 2x
x e − xe + ∫ e dx
2
2
2
1 2 2x 1 2x 1 2x
x e − xe + e + C
2
2
4
∫e
Type D:
ax
sin bxdx or
∫e
ax
Use integration by parts
repeatedly here to find
∫ xe
2x
dx .
cos bxdx
Example 30
Find (a) ∫ e x sin xdx (b) ∫ e ax cos bxdx , where a , b ≠ 0 .
Solution:
∫e
(a)
x
sin xdx
= − ∫ e x d (cos x )
= −e x cos x + ∫ cos xd (e x )
= −e x cos x + ∫ e x cos xdx
= −e x cos x + ∫ e x d (sin x )
= −e x cos x + e x sin x − ∫ sin xd (e x )
= −e x cos x + e x sin x − ∫ e x sin xdx
Rearranging terms, we have
2∫ e x sin xdx = −e x cos x + e x sin x + C
∫e
Note:
x
1
1
1
sin xdx = − e x cos x + e x sin x + C ' , where C ' = C
2
2
2
You can try to find the integral by first expressing ∫ e x sin xdx as ∫ sin xd (e x ) .
You will get the same result at last!
29
∫e
(b)
ax
cos bxdx
1
cos bxd (e ax )
∫
a
1
1
= e ax cos bx − ∫ e ax d (cos bx )
a
a
1
b
= e ax cos bx + ∫ e ax sin bxdx
a
a
1
b
= e ax cos bx + 2 ∫ sin bxd (e ax )
a
a
1
b
b
= e ax cos bx + 2 e ax sin bx − 2 ∫ e ax d (sin bx )
a
a
a
1
b
b2
= e ax cos bx + 2 e ax sin bx − 2 ∫ e ax cos bxdx
a
a
a
Rearranging terms, we have
=
a 2 + b 2 ax
1
b
e cos bxdx = e ax cos bx + 2 e ax sin bx + C
2
∫
a
a
a
a
b
a2
ax
ax
ax
∫ e cos bxdx = a 2 + b 2 e cos bx + a 2 + b 2 e sin bx + C ' , where C ' = a 2 + b 2 C
3.5.3.
Other Examples Involving Integration by Parts
Example 31
(a) Let y = sin −1 x . By considering
dy
1
dx
=
, or otherwise, show that
.
dx
dy
1− x2
(b) Hence, or otherwise, find (i) ∫ sin −1 xdx (ii) ∫ (sin −1 x) 2 dx .
Solution:
(a) From y = sin −1 x , we have x = sin y .
dx
Then, we have
= cos y .
dy
dy  dx 
By inverse function rule of differentiation, we also have
= 
dx  dy 
Also note that cos y = 1 − sin 2 y = 1 − x 2 .
Thus, we have
dy
1
=
.
dx
1− x2
30
−1
=
1
.
cos y
Notes:
y = sin −1 x ,
(1) For
−
π
π
. In this range,
≤ y≤
2
2
cos y is positive, and hence
cos y = 1 − sin 2 y , but not − 1 − sin 2 y .
(2) The result can also be arrived at by finding
∫
1
1− x2
dx using the substitution x = sin θ .
(Refer to Example 19 for details)
(b) (i)
There are two methods for finding this integral.
Method 1:
∫ sin
−1
xdx = x sin −1 x − ∫ xd (sin −1 x)
By
(a),
we
have
d
1
(sin −1 x) =
.
dx
1− x2
x
= x sin x − ∫
−1
dx
1− x2
1
1
= x sin −1 x + ∫
d (1 − x 2 )
2 1− x2
= x sin −1 x + 1 − x 2 + C
Method 2:
Let u = sin −1 x . Then, we have x = sin u , i.e. dx = cos udu .
∫ sin
−1
xdx = ∫ u cos udu
= ∫ ud (sin u )
Reminders:
(1) sin(sin −1 x) = x
(2) We find cos(sin −1 x ) by
= u sin u − ∫ sin udu
= u sin u + cos u + C
constructing the triangle as
shown:
= (sin −1 x ) sin(sin −1 x ) + cos(sin −1 x) + C
= x sin −1 x + 1 − x 2 + C
(ii) Again, there are two methods for this integral.
Method 1:
∫ (sin
−1
x) dx = x(sin x) − ∫ xd (sin
2
−1
2
= x(sin −1 x) 2 − 2 ∫
−1
x)
1
2
sin −1 x
−1
x sin x
dx
1− x2
1− x2
= x(sin −1 x) 2 + 2∫ sin −1 xd ( 1 − x 2 )
= x(sin −1 x) 2 + 2 sin −1 x 1 − x 2 − 2∫ 1 − x 2 d (sin −1 x)
= x(sin −1 x) 2 + 2 sin −1 x 1 − x 2 − 2∫
1− x2
1− x2
= x(sin −1 x) 2 + 2 sin −1 x 1 − x 2 − 2 x + C
31
dx
x
Method 2:
We also let u = sin −1 x . Then, we have x = sin u , i.e. dx = cos udu .
∫ (sin
−1
x) 2 dx = ∫ u 2 cos udu
= ∫ u 2 d (sin u )
= u 2 sin u − ∫ sin ud (u 2 )
= u 2 sin u − 2∫ u sin udu
= u 2 sin u + 2∫ ud (cos u )
= u 2 sin u + 2u cos u − 2∫ cos udu
= u 2 sin u + 2u cos u − 2 sin u + C
= (sin −1 x) 2 sin(sin −1 x) + 2 sin −1 x cos(sin −1 x) − 2 sin(sin −1 x) + C
= x(sin −1 x) 2 + 2 sin −1 x 1 − x 2 − 2 x + C
Example 32
Find (a) ∫ e
sin x
cos xdx for 0 < x < π (b) ∫ sin(ln x) dx for x > 0 .
Solution:
(a) Let u = sin x , then du =
∫e
sin x
1
2 sin x
cos xdx .
cos xdx = ∫ e u ⋅ 2udu
= 2 ∫ ue u du
= 2 ∫ ud (e u )
= 2ue u − 2 ∫ e u du
= 2ue u − 2e u + C
= 2( sin x − 1)e
(b) Let u = ln x , then du =
sin x
+C
1
dx .
x
∫ sin(ln x)dx = ∫ x sin udu
= ∫ e sin udu
u
Repeating the steps in Example 30, we have
1
∫e
u
1
1
sin udu = − e u cos u + e u sin u + C .
2
2
1
cos(ln x ) + e ln x sin(ln x ) + C
2
1
1
= x sin(ln x ) − x cos(ln x) + C
2
2
∫ sin(ln x)dx = − 2 e
ln x
32
Example 33 (Finding
It is given that
∫ tan
p
x sec q xdx for even p and odd q)
∫ sec xdx = ln sec x + tan x + C . (Refer to Example 12 for details)
Find (a) ∫ sec 3 xdx (b) ∫ tan 2 x sec xdx (c) ∫ tan 2 x sec 3 xdx (d) ∫ x 2 x 2 − 1dx .
Solution:
∫ sec
(a)
3
xdx
= ∫ sec x (tan 2 x + 1) dx
= ∫ sec x tan 2 xdx + ∫ sec xdx
= ∫ tan xd (sec x ) + ln sec x + tan x
= sec x tan x − ∫ sec xd (tan x ) + ln sec x + tan x
= sec x tan x + ln sec x + tan x − ∫ sec 3 xdx
Rearranging terms, we have
2∫ sec 3 xdx = sec x tan x + ln sec x + tan x + C1
∫ sec
3
∫ tan
(b)
xdx =
2
1
1
sec x tan x + ln sec x + tan x + C 2
2
2
x sec xdx
= ∫ (sec 2 x − 1) sec xdx
= ∫ sec 3 xdx − ∫ sec xdx
1
1
sec x tan x + ln sec x + tan x − ln sec x + tan x + C 3
2
2
1
1
= sec x tan x − ln sec x + tan x + C 3
2
2
=
Alternatively,
∫ tan
2
x sec xdx
= ∫ tan xd (sec x )
= sec x tan x − ∫ sec xd (tan x )
= sec x tan x − ∫ sec 3 xdx
= sec x tan x − ∫ sec x tan 2 xdx − ∫ sec xdx
= sec x tan x − ln sec x + tan x − ∫ sec x tan 2 xdx
33
Rearranging terms, we have
2∫ tan 2 x sec xdx = sec x tan x − ln sec x + tan x + c
∫ tan
(c)
2
∫ tan
x sec xdx =
2
1
1
sec x tan x − ln sec x + tan x + C3
2
2
x sec 3 xdx
= ∫ (sec 2 x − 1) sec xd (tan x )
= ∫ sec 3 xd (tan x ) − ∫ sec xd (tan x )
= sec 3 x tan x − ∫ tan xd (sec 3 x ) − sec x tan x + ∫ tan xd (sec x )
= sec 3 x tan x − 3∫ tan 2 x sec 3 xdx − sec x tan x + ∫ tan 2 x sec xdx
1
1
= sec 3 x tan x − sec x tan x + sec x tan x − ln sec x + tan x − 3∫ tan 2 x sec 3 xdx
2
2
(by (b))
Rearranging terms, we have
1
1
4 ∫ tan 2 x sec 3 xdx = sec 3 x tan x − sec x tan x − ln sec x + tan x + C 4
2
2
1
1
1
2
3
3
∫ tan x sec xdx = 4 sec x tan x − 8 sec x tan x − 8 ln sec x + tan x + C5
(d) Let x = secθ , then dx = secθ tan θdθ .
∫x
2
x 2 − 1dx = ∫ sec 2 θ sec 2 θ − 1 ⋅ sec θ tan θdθ
= ∫ tan 2 θ sec 3 θdθ
x2 −1
x
1
1
1
sec 3 θ tan θ − sec θ tan θ − ln sec θ + tan θ + C 5
4
8
8
1
1
1
= x 3 x 2 − 1 − x x 2 − 1 − ln x + x 2 − 1 + C 5
4
8
8
=
θ
1
Notes:
(1) The technique of integration by parts followed by rearranging terms is very useful in finding
integrals in the form
∫ tan
p
x sec q xdx , where p is even and q is odd.
(2) Here is a challenging problem: find
∫
x 2 − 4 x + 20dx . Try it out!
[Hint: Make good use of part (a) of Example 33 by using a suitable substitution!]
(For answers, please read the appendix of this note on p.46.)
34
Example 34 (Reduction Formula)
(a) Prove the reduction formulas in Examples 8 and 17, i.e.
(i)
∫ (ln x)
(ii)
∫ sin
n
(b) Let I n = ∫
n
dx = x(ln x ) n − n ∫ (ln x ) n −1 dx for n ≥ 0 ;
n −1
1
xdx = − sin n −1 x cos x +
sin n − 2 xdx for n ≥ 2 .
∫
n
n
dx
, where m and n are positive integers.
( x + 1) n
m
Show that I n +1 =
x
m
mn ( x + 1)
n
+
mn − 1
I n . Hence, or otherwise, find
mn
∫ (x
2
dx
.
+ 1) 3
Solution:
(a) (i)
∫ (ln x)
n
dx
= x(ln x) n − ∫ xd (ln x) n
1
= x(ln x) n − n ∫ x(ln x) n −1 ⋅ dx
x
= x(ln x) n − n ∫ (ln x) n −1 dx
(ii)
∫ sin
n
xdx
= − ∫ sin n −1 xd (cos x )
= − sin n−1 x cos x + ∫ cos xd (sin n −1 x )
= − sin n−1 x cos x + ( n − 1) ∫ sin n − 2 x cos 2 xdx
= − sin n−1 x cos x + ( n − 1) ∫ sin n − 2 x(1 − sin 2 x ) dx
= − sin n−1 x cos x + ( n − 1) ∫ sin n − 2 xdx − ( n − 1) ∫ sin n xdx
Rearranging terms, we have
(n − 1 + 1) ∫ sin n xdx = − sin n−1 x cos x + (n − 1) ∫ sin n−2 xdx
∫ sin
n
1
n −1
xdx = − sin n−1 x cos x +
sin n− 2 xdx
∫
n
n
Notes:
(1) Rearranging terms is a very commonly used technique used in proving reduction formula using
integration by parts.
(2) In proving reduction formulas involving trigonometric functions, the trigonometric identities
sin 2 x + cos 2 x = 1 , tan 2 x + 1 = sec 2 x and cot 2 x + 1 = csc 2 x are helpful on most occasions.
35
(b)
In
=


x
1
− ∫ xd  m
n
n 
( x + 1)
 ( x + 1) 
m
x
x (mx m −1)
= m
+ n∫ m
dx
( x + 1) n
( x + 1) n +1
x
xm
= m
+ mn ∫ m
dx
( x + 1) n
( x + 1) n +1
x
xm +1
1
= m
+ mn ∫ m
dx − mn ∫ m
dx
n
n +1
( x + 1)
( x + 1)
( x + 1) n +1
x
= m
+ mnI n − mnI n +1
( x + 1) n
Rearranging terms, we have
x
+ (mn − 1) I n
( x + 1) n
x
mn − 1
=
+
In
m
n
mn
mn( x + 1)
mnI n+1 =
I n+1
m
First, we take m = 2 and n = 2 .
dx
x
( 2)( 2) − 1
dx
∫ ( x 2 + 1) 3 = (2)(2)( x 2 + 1) 2 + (2)(2) ∫ ( x 2 + 1) 2
∫ (x
For
∫ (x
For
2
2
dx
, take m = 2 and n = 1 .
+ 1) 2
dx
x
3
x
( 2)(1) − 1 dx 
=
+ 
+

3
2
2
2
4  ( 2)(1)( x + 1)
( 2)(1) ∫ x 2 + 1 
+ 1)
4( x + 1)
x
3x
3
dx
=
+
+ ∫ 2
2
2
2
4( x + 1)
8( x + 1) 8 x + 1
∫x
dx
, let x = tan θ , then dx = sec 2 θdθ .
+1
2
dx
sec 2 θdθ
∫ x 2 + 1 = ∫ tan 2 θ + 1
= ∫ dθ
= θ + C1
= tan −1 x + C1
Thus, we have
∫ (x
Note: The integral
2
dx
x
3x
3
=
+
+ tan −1 x + C .
3
2
2
2
+ 1)
4( x + 1)
8( x + 1) 8
∫ (x
dx
can also be found by letting x = tan θ at the very beginning.
+ 1) 3
2
The reduction formula can even be found by differentiation. Try them out!
36
3.6 Integration of Miscellaneous Functions
Some functions can be integrated with a combination of the above techniques.
3.6.1.
Finding Integrals in the Form
Ax 2 + Bx + C
∫ Dx 2 + Ex + F dx
We follow the following steps:
(In the following, all capital letters represent constants.)
Ax 2 + Bx + C
∫ Dx 2 + Ex + F dx as
Hx + I
(2) Express ∫
dx as
2
Dx + Ex + F
(1) Express
(3) Find
∫ Gdx , ∫ Dx
2
∫ Gdx + ∫ Dx
∫ Dx
2
Hx + I
dx .
+ Ex + F
2
J
K
d ( Dx 2 + Ex + F ) + ∫
dx .
2
+ Ex + F
Dx + Ex + F
J
d ( Dx 2 + Ex + F ) and
+ Ex + F
∫ Dx
2
K
dx one by one.
+ Ex + F
The integral is then found.
Example 35
Find
3x 2 − 4 x − 5
∫ x 2 + 2 x + 5 dx .
Solution:
This step can be replaced by
3x 2 − 4 x − 5
3( x 2 + 2 x + 5)
10 x + 20
dx
=
dx
−
dx
∫ x 2 + 2x + 5
∫ x 2 + 2x + 5
∫ x 2 + 2x + 5
long division.
5( 2 x + 2)
1
= ∫ 3dx − ∫ 2
dx − 10 ∫ 2
dx
x + 2x + 5
x + 2x + 5
1
1
= 3 x − 5∫ 2
d ( x 2 + 2 x + 5) − 10 ∫ 2
dx
x 2 + 2x + 5
( x + 2 x + 1) + 4
x + 2x + 5
= ( x + 1) 2 + 4
1
2
= 3 x − 5 ln( x + 2 x + 5) − 10 ∫
dx
> 0 for all real number x
( x + 1) 2 + 4
Hence, absolute sign is
1
2
not required here for
For ∫
dx , let x + 1 = 2 tan θ , then dx = 2 sec θdθ .
( x + 1) 2 + 4
ln( x 2 + 2 x + 5) .
1
2 sec 2 θ
dx
=
∫ ( x + 1) 2 + 4
∫ (2 tan θ ) 2 + 4 dθ
1
= ∫ dθ
2
1
= θ + C1
2
1
x +1
= tan −1
+ C1
2
2
Thus, we have
3x 2 − 4 x − 5
2
−1 x + 1
∫ x 2 + 2 x + 5 dx = 3x − 5 ln( x + 2 x + 5) − 5 tan 2 + C .
37
Example 36
1
A
Bx + C
, where A, B and C are constants. Find A, B and C.
≡
+ 2
x +1 x +1 x − x +1
(a) Let
3
(b) Hence, find
∫x
1
dx .
+1
3
Solution:
(a) From the question, we have
1
A
Bx + C
.
≡
+ 2
x +1 x +1 x − x +1
3
1
A( x 2 − x + 1) + ( Bx + C )( x + 1)
≡
x3 +1
( x + 1)( x 2 − x + 1)
( A + B) x 2 + ( B + C − A) x + ( A + C )
x3 + 1
Hence, we have 1 ≡ ( A + B ) x 2 + ( B + C − A) x + ( A + C ) .
≡
A + B = 0

Comparing the coefficients of like terms, we have B + C − A = 0 .
A + C = 1

1
1
2
Solving the above system of linear equations, we have A = , B = − and C = .
3
3
3
(b)
∫x
1
1 1
1
x−2
dx = ∫
dx − ∫ 2
dx
+1
3 x +1
3 x − x +1
1 1
1
2x − 4
= ∫
d ( x + 1) − ∫ 2
dx
3 x +1
6 x − x +1
1
1
2x − 1
1
3
dx + ∫
dx
= ln x + 1 − ∫ 2
2
2
3
6 x − x +1
6  2
1
1 
x − x +    + 1 −  
 2  
 2

3
1
1
1
1
1
d ( x 2 − x + 1) + ∫
dx
= ln x + 1 − ∫ 2
2
3
6 x − x +1
2 
1
3
x−  +
2
4

For
∫
∫
1
2
1
3
x−  +
2
4

1
2
∫x
1
3
3
=
tan θ , then dx =
sec 2 θdθ .
2
2
2
 2x − 1 
3
sec 2 θ
2
2
2
dθ =
dθ =
tan −1 
 + C1
θ + C1 =
∫
2
∫
2  3
3
3
3
 3 

3


 2 tan θ  + 4


dx =
1
3
x −  +
2
4

Thus, we have
dx , let x −
 2x − 1 
1
1
1
1
tan −1 
dx = ln x + 1 − ln( x 2 − x + 1) +
 + C .
3
6
+1
3
 3 
3
38
3.6.2.
General t-substitution for integrals involving trigonometric functions
There are some integrals that involve trigonometric functions in the denominator and/or numerator
in the integrand, e.g.
1
∫ 1 + sin x dx
and
1
∫ sin x + cos x + 1 dx , etc. To find these integrals, we can
x
use a special method which makes the whole integral be in terms of the variable t = tan .
2
Before talking about how we can use this method, we first see how we can express the three basic
trigonometric functions, sin x , cos x and tan x , in terms of tan
x
.
2
Example 37
x
Let t = tan . Express sin x , cos x and tan x in terms of t.
2
Solution:
x
x
sin x = 2 sin cos
2
2
x
sin
2 cos 2 x
=2
x
2
cos
2
x
2 tan
2
=
x
sec 2
2
x
2 tan
2
=
x
1 + tan 2
2
2t
=
1+ t2
2 tan
x
cos x = 2 cos 2 − 1
2
2
=
−1
2 x
sec
2
2
=
−1
2 x
1 + tan
2
2 − (1 + t 2 )
=
1+ t 2
1− t2
=
1+ t 2
tan x =
x
2
1 − tan 2
=
x
2
2t
1− t2
Notes:
You may also consider constructing a right-angled triangle as shown after finding tan x . In this
case, sin x and cos x can also be found. However, this is NOT considered a good method, as you
can only find sin x and cos x for 0 < x < π in this case. Using trigonometric identities would
be a better method as the result is true for all real values of x.
There are also other methods to find out the way to express the three functions
in terms of t. Interested readers can search for more information on the internet.
39
Example 38
Let I = ∫
1
x
dx . Using the substitution t = tan , or otherwise, find I.
1 + sin x
2
Solution:
x
Let t = tan .
2
1
x
sec 2 dx
2
2
2
dx =
dt
2 x
1 + tan
2
2
=
dt
1+ t2
dt =
1
dx
1 + sin x
1
2
=∫
⋅
dt
2t 1 + t 2
1+
1+ t2
2
=∫ 2
dt
t + 2t + 1
1
= 2∫
d (t + 1)
(t + 1) 2
2
=−
+C
t +1
2
=−
+C
x
tan + 1
2
I =∫
(by Example 37)
Notes:
(1) Readers should pay attention to the way we express dx in terms of t.
(2) This integral can be especially found in an easier way:
x
sec 2
1
1
2 dx
I =∫
dx = ∫
dx = ∫
x
x
x
x
x
x
x
sin 2 + 2 sin cos + cos 2
(sin + cos ) 2
(tan + 1) 2
2
2
2
2
2
2
2
2
x
2
=∫
d (tan + 1) = −
+C
x
x
2
2
(tan + 1)
tan + 1
2
2
(3) You may also consider:
1
1 − sin x
1 − sin x
d (cos x )
dx = ∫
dx = ∫ sec 2 xdx + ∫
= tan x − sec x + C '
2
2
x
cos x
cos 2 x
∫ 1 + sin x dx = ∫ 1 − sin
*Think about*:
Can you show that the two answers obtained differ by a constant (1, in this case)?
40
Example 39
Find
1
∫ 2 sin x − 3 cos x + 5 dx .
Solution:
2
x
Again, we let t = tan , then dx =
dt . (Refer to Example 38)
2
1+ t2
1
1
2
⋅
dt
2
1+ t2
 2t   1 − t 
+5
2
− 3
2 
2 
1+ t  1+ t 
2
=∫
dt
4t − 3 + 3t 2 + 5 + 5t 2
1
=∫ 2
dt
4t + 2t + 1
1
=∫
dt
1
1
 2 1
4 t + t +  + 1 −
2 16 
4

1
=∫
dt
2
3
 1
4 t +  +
4
 4
1
1
= ∫
dt
4  1 2 3
t +  +
 4  16
∫ 2 sin x − 3 cos x + 5 dx = ∫
Let t +
1
3
3
=
tan θ , then dt =
sec 2 θdθ .
4
4
4
1
1 3
sec 2 θ
dx
=
⋅
dθ
2
∫ 2 sin x − 3 cos x + 5
4 4 ∫ 3

3


 4 tan θ  + 16


1
=
∫ dθ
3
1
=
θ +C
3
1
4t + 1
=
tan −1
+C
3
3
x
4 tan + 1
1
2
=
tan −1
+C
3
3
Notes:
It is common that we have to use a further substitution to find the integral obtained by letting
x
t = tan , and the techniques in Section 3.6.1 are useful in most situations.
2
41
4.
Applications of Indefinite Integration
Indefinite integration is useful in both mathematics and daily life applications.
A.
Applications in Coordinate Geometry
It is known that f ' ( x ) represents the slope of the tangent of a point (x,y) to the curve y = f (x ) .
By integrating f ' ( x ) with respect to x, it is expected that we can obtain the original curve
y = f (x ) .
However, is it the case? No!
Integrating f ' ( x ) w.r.t. x, we have
∫ f ' ( x)dx = y + C =
y
f ( x) + C
We observe that there is an extra constant C. In fact, what
we have obtained is a family of curves, which can be
obtained by translating the curve of y = f (x ) upwards or
downwards.
Slope = f ' ( x )
We can further observe that all these curves have the same
f ' ( x ) , i.e. the slope of tangent to all curves at point (x,y) is
the same.
In order to find out the equation of a specific curve instead
of a family of curves, we need the coordinates of a specific
point to find out the constant C. This restricts the possibility
of curves to only one curve, but not the family of curves as
x
O
Figure 3 The slope of tangent to the
curve at point (x,y) is equal to f ' ( x ) .
y
mentioned.
Similarly, suppose f ' ' ( x ) is given, we have
∫ f ' ' ( x)dx =
f ' ( x ) + C1
∫ [ f ' ( x) + C ]dx =
1
x
O
f ( x ) + C1 x + C 2
To find out f (x ) from f ' ' ( x ) , we need to integrate twice.
Figure 4 These curves, obtained by
translating y = f (x ) upwards or
downwards, have the same f ' ( x ) .
We also need the coordinates of two points on the curve.
By substituting them into y = f ( x) + C1 x + C 2 , we can find out the constants C1 and C2 .
Finally, we can find out the original function f (x ) .
Refer to the examples on the next page.
42
Example 40
The slope of any point (x,y) on a curve Q is
x
1+ x2
. It is known that Q passes through the origin.
Find the equation of Q.
Solution:
dy
x
=
.
dx
1+ x2
Integrating both sides w.r.t. x, we have
From the question, we have
y=∫
x
dx
1+ x2
1
1
= ∫
d (1 + x 2 )
2
2 1+ x
= 1+ x2 + C
Put (0,0) into y = 1 + x 2 + C , we have
0 = 1 + 02 + C
C = −1
Thus, the equation of Q is y = 1 + x 2 − 1 .
Example 41
For curve Y, it is given that
d2y
= e 2 x . If Y passes through points A(0,1) and B(1,1), find the
2
dx
equation of curve Y.
Solution:
d2y
dy
1
= e 2 x , by integration, we have
= ∫ e 2 x dx = e 2 x + C1 .
2
dx
2
dx
1
1

By further integration, we have y = ∫  e 2 x + C1 dx = e 2 x + C1 x + C 2
4
2

From
Put (0,1) and (1,1) into y =
1 2x
3
1
and C1 + C 2 = 1 − e 2 .
e + C1 x + C 2 , we get C 2 =
4
4
4
Solving the above system of linear equations, we get C1 =
Thus, the equation of curve Y is y =
1 2x 1 − e2
3
e +
x+ .
4
4
4
43
1− e2
3
and C 2 = .
4
4
B. Applications in Physics
Consider an object travelling in a straight line:
Q
P
We define some terms as follows:
Displacement (s) is the distance of an object from its original position, with the consideration of
direction.
Velocity (v) is the derivative of displacement with respect to time (t).
Acceleration (a) is the derivative of velocity with respect to time.
If we take right hand side as positive direction, If PQ = 3 m, then the displacement of P is 3 m. On
the contrary, if we take left hand side as positive direction, then the displacement of P is −3 m.
dv d 2 s
=
.
dt dt 2
By integrating a or v with respect to t, we can obtain v and s respectively.
Note that a =
Example 42
The velocity of a particle travelling along a straight line is given by v(t ) = 5 − 2t , where v(t ) is the
velocity (in m/s) and t is time (in second). It is known that the particle passes through its original
position after 4 seconds. Take right hand side as positive direction.
(a) Find the displacement of the particle after t seconds.
(b) What is the displacement of the particle when it is momentarily at rest?
Solution:
(a) Let the displacement of the particle after t seconds be s (t ) m.
Then s(t ) = ∫ v(t ) dt = ∫ (5 − 2t ) dt = 5t − t 2 + C .
When t = 4 , s(t ) = 0 .
5( 4) − (4) 2 + C = 0
C = −4
Thus, the displacement of the particle is ( − t 2 + 5t − 4 ) m.
(b) The particle is momentarily at rest when v (t ) = 0 , i.e. t =
2
9
5
5
The displacement of particle = −   + 5  − 4 = m
4
2
2
44
5
.
2
C. Other Applications
Other than in physics, in many situations, if the rate of change is given, we can find out the original
quantity by integrating the rate with respect to time.
Example 43
The figure on the right shows an inverted circular cone with base radius 30
cm and height 40 cm. At the beginning, the cone was full of water. A small
hole was then drilled at the apex of the cone and water is flowing out of
the cone at a rate of ( 2t + 1) cm3/s, where t is time in second.
(a) Express the volume of water remaining in the cone after t seconds.
(b) After how many seconds will the cone be empty?
Correct your answer to the nearest second.
Figure 5 An inverted circular cone
Solution:
(a) Let the volume of water remaining in the cone be V cm3.
Then, we have
dV
Negative sign is added
= −( 2t + 1)
dt
since the volume of
V = − ∫ ( 2t + 1) dt
water is decreasing.
2
= −t − t + C
1
When t = 0 , we have V = π (30) 2 ( 40) = 12000π cm 3 .
3
Thus, the volume of water remaining in the cone after t seconds is (12000π − t − t 2 ) cm3.
(b) When the cone is empty, V = 0 .
12000π − t − t 2 = 0
t 2 +t − 12000π = 0
t=
− 1 ± (1) 2 − 4(1)(−12000π )
2(1)
− 1 + 48000π + 1 − 1 − 48000π + 1
or
(rejected,Q t > 0)
2
2
≈ 194
=
Thus, the required time is 194 seconds.
More applications of integration will be found in the topic ‘Definite integrals’.
The writer would like to thank Mr. Yue Kwok Choy for his kind reading and checking of this
article.
-END45
Appendix 1
Proof of Method of Substitution in Integration
To prove:
Let u = g ( x ) . Then, we have
Consider the function
Then
∫ f ( g ( x)) g ' ( x)dx = ∫ f (u )du .
y = F ( g ( x )) , where F ( g ( x )) is a primitive function of f ( g ( x )) .
dy
= F ' ( g ( x )) g ' ( x ) = f ( g ( x )) g ' ( x ) .
dx
Integrating both sides w.r.t. x, we have y = ∫ f ( g ( x)) g ' ( x)dx .
But
dy
= F ' (u ) = f (u ) .
du
Integrating both sides w.r.t. u, we have y = ∫ f (u ) du .
Then, we have
∫ f ( g ( x)) g ' ( x)dx = ∫ f (u )du .
This completes this proof.
Appendix 2
Finding ∫ x 2 − 4 x + 20dx
∫
x 2 − 4 x + 20dx = ∫ ( x − 2) 2 + 4 2 dx
Now let x − 2 = 4 tan θ , then dx = 4 sec 2 θdθ .
∫
x 2 − 4 x + 20dx = ∫ ( x − 2) 2 + 4 2 dx
= ∫ (4 tan θ ) 2 + 4 2 ⋅ 4 sec 2 θdθ
= 16∫ sec 3 θdθ
= 8 sec x tan x + 8 ln sec x + tan x + C (by Example 33(a))
 x 2 − 4 x + 20  x − 2 
x 2 − 4 x + 20 x − 2

= 8
+
+C
 + 8 ln

 4 
4
4
4


1
= ( x − 2) x 2 − 4 x + 20 + 8 ln x 2 − 4 x + 20 + x − 2 + C'
2
46