Download Handout 3 – Integration

Integration
October 25, 2016
1
Introduction
We have learned in previous chapter on how to do the differentiation. It is a conventional in mathematics that we are supposed to learn about the integration as well. As
you may have already know, integration is the inverse of differentiation, or sometimes
called antiderivative. In fact, by realising the relationship between these two, has laid
out the foundation of what we now called Calculus which was invented by Newton
and Leibniz in 17th century.
Our main goal for this chapter is to learn about the integration of the transcendental functions. First we will start on reviewing the integration of simple integrand(the
equation that we want to integrate), the indefinite and the definite integral. Next we
will learn the three techniques of integration, and then we will apply these techniques
on integrating transcendental functions accordingly.
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Review of Integration
2.1
Basic Integral Formulae
Example 1 : Evaluate the following indefinite integrals
Z
1.
x3 dx
Z
√
x dx
Z
dx
x3
Z
dx
√
x
2.
3.
4.
1
Z
5.
sin x dx
Z
6.
cos x dx
Z
7.
Z
8.
Z
9.
Z
10.
Z
Z
12.
x6 dx
dx
x8
4
1
1
x3
Z
13.
Z
14.
Z
15.
Z
16.
dx
5
x− 4 dx
sec2 x dx
cosecx dx
(1 + tan2 x) dx
Z
dx
sec x
Z
sin x
cos2 x
Z
sin 2x
2 cos x
17.
18.
2.2
sech2 x dx
x 5 dx
11.
19.
ex dx
Integrals with Sum, Difference and Factor
Example 2 : Evaluate the following indefinite integrals
Z
1. (3x2 + 8) dx
2
Z
2.
(−4 sech x tanh x) dx
Z
(x4 + 2x − 6) dx
3.
√
4
(3x2 + 3 4 x − 3 ) dx
x
Z
4.
Z
(4 + tan2 x) dx
5.
Z
6.
(sin
x
x
− cos ) dx
2
2
Z
(x + 1)(3x − 2) dx
7.
1 + ex sec x
dx
sec x
Z
8.
2.3
Definite Integral
Example 3: Evaluate the following integrals
Z 2
(3x2 + 7) dx
1.
0
2
Z
(x + 2)(x + 3) dx
2.
1
2
Z
3.
1
3
3.1
2x3 + 1
dx
x3
Techniques of Integration
Integration by Substitution
Method of substitution is a straightforward from chain rule, and this method is what
you should consider first before other methods. We usually choose the substitution
for the ‘inside’ function and then manipulate some algebra for what’s left on the
’outside’ function. Example 4: Evaluate the following integral by using the proper
substitution
Z
1. (4x + 1)2 dx
3
Z
2.
Z
3.
Z
4.
c dx
ax + b
e4x dx
e−3x+5 dx
Z
5
(10x − 9) 2 dx
5.
Z
6.
sin 15x dx
Z
7.
cos(2x + 3) dx
Z
tanh(3 − 2x) dx
8.
Z
9.
x2
2x + 3
dx
+ 3x + 2
x2
x
dx
−4
Z
10.
Z
12x2 + 16
dx
x3 + 4x
Z
x + cos x
dx
3x2 + 6 sin x
11.
12.
Z
13.
tan x dx
Z
14.
sin x
dx
1 − cos x
Z
15.
sec x dx
Z
16.
Z
17.
Z
18.
4x(2x2 − 3)6 dx
sin x cos4 x dx
ex (ex − 1)3 dx
4
Z
ln x
dx
x
Z
dx
x ln x
19.
20.
Z
21.
Z
22.
Z
23.
Z
24.
Z
25.
Z
26.
Z
27.
Z
28.
Z
29.
2
xe−x dx
2x(x + 2)5 dx
√
x 3x − 2 dx
√
x
dx
x+4
2x3 cos x4 dx
ex sin ex dx
etan x secx dx
5x cos(5 − x2 ) dx
Z
dx
√
1+ x
Z
dx
1 + ex
Z
x
dx
x−4
30.
31.
32.
Z
33.
x sech2 (2x2 + 3) tanh(2x2 + 3) dx
sin2 (6x + 7) dx
Sometimes the substitution is given to us. Have in mind that the substitution will
help us to simplify the integrand, not making it more difficult to integrate, which is
the sole reason on why we are learning to find the appropriate substitution in using
this technique.
Example 5 Evaluate the following integrals by using the given substitution
5
0
Z
1.
√
3x 1 − 3x dx with substitution u2 = 1 − 3x.
−1
1
2
Z
2.
√
0
2
Z
3.
√
4x
dx with substitution x = sin θ.
1 − x2
4 − x2 dx with substitution x = 2 sin θ.
−2
3.2
Integration by Parts
Integration by parts is a powerful method to evaluate integrals when the integrand is
in the form of product of algebraic and transcendental, such as
Z
Z
Z
x
x ln x dx,
xe dx,
ex cos x dx.
General formula for integration by parts is
Z
u dv = uv − v du
Tips
Z
• If the integral is in the form of
Z
• If the integral is in the form of
xn ln xdx, take u = ln x and dv = xn dx.
Z
ax
eax cos bx dx, take u = sin bx
e sin bx dx or
and dv = eax dx. Or, we can use the substitution the other way around by taking u = eax and dv = sin ax dx.
Z
• If the integral is in the form of
n ax
x e dx,
Z
n
Z
x sin ax dx or
xn cos ax dx,
use the substitution u = xn and dv = eax dx, sin ax dx, cos ax dx.
Example 6 Evaluate the following integral using the following integrals
Z
1.
x ln x dx
Z
2.
ln x dx
Z
3.
Z
4.
xex dx
1
xe−x dx
0
6
Z
5.
x sin x dx
π
2
Z
6.
x cos x dx
0
Z
x2 cos x dx
7.
π
2
Z
8.
x2 sin x dx
0
Z
9.
Z
10.
3.3
ex cos 2x dx
x2 ln x dx
Tabular Method
When integrating by parts have many repeated integrations and differentiations, tabular method is a very useful trick since it’s making the integration by part procedure
more neat.
Example 7 Evaluate the integral using the tabular method
Z
1.
x3 e2x dx.
Z
2.
Z
3.
Z
4.
Z
5.
x2 sin 3x dx.
x sec2 x dx.
3
x2 (x − 2) 2 dx.
e3x cos 2x dx
Z
6.
cos 5x sin 4x dx
7
3.4
Integration by Partial Fraction
By using this technique, we rewrite the integral which initally in product and quotient
form, in terms of fraction. In practice the integral is in the form of polynomial of
rational functions, where the denominator usually can be factorised. Though pretty
much most of polynomial of degree 2 can be factorised, bear in mind that if the
factor on the numerator results to a complicated algebra, you are advised to use
other method ie substitution or inverse trigonometric and inverse hyperbolic.
If the numerator has the same order of degree or more with the denominator, it
becomes an improper fraction. For this particular case, you are required to use long
division of polynomials.
Example 8 Evaluate the integral by using partial fractions
Z
3x + 2
dx
1.
2
x + 3x + 2
Z
x−3
2.
dx
3x2 + 2x − 5
Z
6x + 7
dx
3.
(x + 2)2
Z
5x2 + 20x + 6
4.
dx
x3 + 2x2 + x
Z
x
5.
dx
x+1
Z
x3
dx
6.
x−1
Z
x2
7.
dx
x2 + 3x + 2
Z
2x5 − 5x
8.
dx
(x2 + 2)2
Rational Functions of sin x and cos x
When the integral is in the form
1
,
a cos x + b sin x + c
it can be integrated by using the substitution
t = tan
8
x
2
. We can show that
sin x =
2t
,
1 + t2
cos x =
1 − t2
,
1 + t2
tan x =
2t
,
1 − t2
dx =
2dt
.
1 + t2
We can also find another identity that come in handy by using the substitution.
tan
sin x2
2 sin x2 cos x2
x
sin x
=
.
=
=
x
x
2
2
cos 2
2 cos 2
1 + cos x
It is encouraged for you to memorise only on sin x formula above, and then draw the
dx
, this term is obtained
triangle to obtain the terms cos x and tan x easily. As for
dt
x
by differentiating the given substitution tan = t.
2
Example 9 Evaluate the following integrals
Z
dx
1.
cos x − sin x − 1
Z
dx
2.
5 cos x − 3
Z π
2
dx
3.
0 3 + 5 sin x
Z π
3
2 dx
4.
using substitution tan x = t.
π
sin
2x
6
4
Integrals of Hyperbolic Functions
Integrals of hyperbolic functions are similar to the integrals of trigonometric functions.
You may refer to the formula to find the antiderivative of the hyperbolic functions.
As for the more complicated integrals, we can use the three methods described earlier.
Example 10
1. Evaluate the following integrals
Z
a)
sinh 2x dx
Z
b)
5 cosh 3x dx
Z
c)
sinh2 x dx
Z
d)
cosh2 2x dx
9
Z
e)
Z
f)
3 sech2 4x dx
cosh2 (5x + 2) dx
2. By using the definitions show that
1
= cosh 2x − sinh 2x.
cosh 2x + sinh 2x
Hence show that
Z
0
1
1
1
dx = (1 − e−2 ).
cosh 2x + sinh 2x
2
3. By using the tabular method, evaluate the following integral
Z
a)
x3 cosh 2x dx
Z
b)
x2 sinh 3x dx
Z
c)
e3x cosh 2x dx
Z
d)
sin 2x sinh 3x dx
5
Integration Involving Inverse Trigonometric Functions
In previous chapter we have learned how to do the differentiation of the inverse
function using the formula. Hence to find the integral, the idea is the same. What
we need to do is to manipulate the integrand(provided that we are sure that the
integral will result to inverse trigonometric or inverse hyperbolic), so that it matches
the formula. Recall that the differentiation of inverse trigonometric is given as
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d
[sin-1 x] =
dx
d
[cos-1 x] =
dx
d
[tan-1 x] =
dx
d
[cot-1 x] =
dx
d
[sec-1 x] =
dx
d
[cosec-1 x] =
dx
1
1 − x2
−1
√
1 − x2
1
1 + x2
−1
1 + x2
1
√
|x| x2 − 1
−1
√
.
|x| x2 − 1
√
Therefore the integral formula is the antiderivative of those and given as
Z
1
√
dx = sin-1 x + C, |x| < 1
2
1−x
Z
−1
√
dx = cos-1 x + C, |x| < 1
2
Z 1−x
1
dx = tan-1 x + C
1 + x2
Z
−1
dx = cot-1 x + C
1 + x2
Z
1
√
dx = sec-1 x + C, |x| > 1
2
|x| x − 1
Z
−1
√
dx = cosec-1 x + C, |x| > 1
2
|x| x − 1
Generally, the integral for the inverse trigonometric will look like this
Z
dx
√
,
2
a − x2
and at first glance, we know it looks like the differentiation equation of
we let x = au, then dx = a du. Hence
Z
Z
x
du
dx
√
√
=
= sin-1
+ C.
a
a2 − x 2
a2 − u2
d
[sin-1 x]. If
dx
The same procedure applied on the appropriate integrand similar to the form of
differentiation of tan-1 x and sec-1 x. The general formula for this type of integrands
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are given as
Z
1
-1 x
√
+ C, |x| < 1
dx = sin
2 − x2
a
a
Z
1
1
-1 x
+C
dx = tan
a2 + x 2
a
a
Z
x
1
1
√
+ C, |x| > 1,
dx = sec-1
a
a
|x| x2 − a2
where the integration formula that associates with cos-1 x, cot-1 x and cosec-1 x are
just the negative terms of the above integrands respectively.
It would be useful if we familiarise ourselves with the technique of completing the
square that will be used frequently in transforming the integrand. Completing the
square is transforming the quadratic equation
ax2 + bx + c = 0,
into
a(x + d)2 − e = 0,
b
b2
where d =
and e = c − .
2a
4a
Example 11 Evaluate the following integrals.
Z
dx
√
1.
25 − x2
Z
dx
2.
9 + x2
Z
dx
√
3.
|x| x2 − 4
Z
dx
√
4.
1 − 9x2
Z
dx
5.
4 + 9x2
Z
dx
√
6.
|x| 4x2 − 5
Z
dx
√
7.
3 + 2x − x2
Z
dx
8.
2
2x + 4x + 5
12
Z
dx
√
|x + 1| x2 + 2x
9.
1/4
Z
√
10.
0
2
Z
11.
x2
1
5
Z
12.
3
dx
1 − 16x2
dx
− 2x + 2
dx
√
(x − 1) x2 − 2x − 3
1
Z
sin-1 x dx
13.
1
2
√
3
2
Z
14.
cos-1 x dx
1
2
Z
15.
1
x tan-1 x dx
0
6
Integration Involving Inverse Hyperbolic Functions
The procedure of solving the integration involving inverse hyperbolic functions is
similar to the procedure of solving the integrating involving inverse trigonometric
functions. The general formula for the integrand that will result to the inverse hyperbolic functions are given as
Z
x
dx
-1
√
= sinh x
+ C, a > 0
a
a2 + x 2
Z
x
dx
-1
√
= cosh x
+ C, x > a
2
2
a
Z x −a
dx
1
-1 x
=
tanh
+ C, if |x| < a
a2 − x 2
a
a
Z
dx
1
-1 x
=
coth
+ C, if |x| > a
a2 − x 2
a
a
Z
dx
1
-1 x
√
= − sech
+ C, if 0 < x < a
2 − x2
a
a
x
a
Z
dx
1
-1 x
√
= − cosech
+ C, if 0 < x < a
a
a
x a2 + x 2
13
Example 12 Evaluate the following integral.
Z
dx
√
1.
x 1 + x2
Z
ex
2.
16 − e2x
Z
dx
√
3.
25 + x2
Z
dx
√
4.
, x>4
2
x − 16
Z
dx
5.
, |x| < 2
4 − x2
Z 1
dx
√
6.
9 + 4x2
0
Z 5
dx
√
7.
4x2 − 36
4
Z 2
dx
8.
2
0 25 − 4x
Z 4
dx
√
9.
2
2x − 4x + 5
1
Z 5
2
dx
√
10.
3
x2 − x
2
Z 1
dx
√
11.
2
0 (x + 1) x + 1
Z
x+1
√
12.
dx
x2 + 1
Z √3
2
13.
x2 tanh-1 x dx
1
2
7
Conclusion
In this topic, you are required to know how to integrate the integrals that involve
trigonometric, inverse trigonometric, hyperbolic and inverse hyperbolic functions.
What you should know first is to master all the techniques of integration which
are
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• Integration by Substition
• Integration by Parts( and Tabular Methods)
• Integration by Partial Fractions
Once you have familiar yourselves with all these techniques, the next step is to recognised the pattern of which technique you should use given an integral. There is no
trick or shortcut to it. It is advised for you to do as many integration practises as
you can, then you will be able to recognised the pattern.
However it is suggested that (but don’t take the following as a strict rule that
you have to follow) for you to always try the method of substitution first. If it’s not
working ie the resulting substitution is not a simple algebra, then you can try partial
fraction (if the integral is in fraction), or by integration by parts.
If you were given an integration in the form of fraction with the denominator is
in square root form, try the method of substitution first, if it isn’t working, see if the
integration of inverse trigonometric or hyperbolic works.
If you were given an integration in a form of fraction of polynomial, check the
order of the power. If the order of the numerator is the same or greater than the
denominator, then use long division first.
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