Download Homework 4. Techniques of Anti

5
Techniques of Anti-differentiation
5.1
Differential Notation 1
Calculate the differential of the following functions by using the definition
dy = y 0 (x)dx.
Express the result in terms of the product between y 0 (x) and the differential of x, dx.
For example, given
y(x) = 3x + sin(2x),
its differential is
dy = y 0 (x)dx = (3 + 2 cos(2x))dx.
(a) f (x) = ex
2
(d) f (x) = arcsin(x)
(g) y = x6 + 2x4 − 2x
√
(b) f (x) = (x + 1)2
(c) f (x) =
(e) f (x) = x2 + 3x + 1
(h) y = (x − 2)2 (x + 1)5
(f) f (x) = cos(2x)
(i) y = x/(x + 3)
x
Solution
2
2
(a) d(ex ) = 2xex dx
(b) d((x + 1)2 ) = 2(x + 1) dx
√
(c) d( x) = (1/2)x−1/2 dx
1
dx
1 − x2
(f) d(cos(2x)) = −2 sin(2x) dx
(e) d(x2 + 3x + 1) = (2x + 3) dx
(g) dy = (6x5 + 8x3 − 2) dx
(h) dy = (2(x − 2)(x + 1)5 + 5(x − 2)2 (x + 1)4 ) dx
(i) dy = ((1(x + 3) − x)/(x + 3)2 ) dx = (3/(x + 3)2 ) dx
5.2
(d) d(arcsin(x)) = √
Differential Notation 2
Given the differential of a function y in terms of the product between its derivative y 0 (x) and
the differential of x, (dx), express the same differential in terms of the the differential of y itself,
i.e., dy. Then, use the result combined with the fundamental theorem of calculus to calculate
the corresponding indefinite integrals.
For example, if we know that the differential of a function y(x) is given by
(3 + 2 sin(2x))dx =?,
(Or equivalently,
Z
(3 + 2 sin(2x))dx =?)
we now need to figure out that y(x) = 3x − cos(2x) and express the differential given above in
terms of the differential of y itself
(3 + 2 sin(2x))dx = d(3x − cos(2x)).
1
Using this result and the fundamental theorem of calculus, we can solve the following
Z
(3 + 2 sin(2x))dx =
Z
d(3x − cos(2x)) = 3x − cos(2x) + C.
(Those marked by a star are a little bit more difficult since they involve composite functions.)
R 3
R 3/2
(a) x3 dx =?,
x dx =?
(b) x3/2 dx =?,
x dx =?
√
√
R
R
(d) xdx =?, R xdx =?
(c) x13 dx =?, R x13 dx =?
1
1
√
√
dx =?
(f) e−2x dx =?,
e−2x dx =?
(e) x dx =?,
x R
R
1
1
(g) (x + 1)2 dx =?,
(x + 1)2 dx =? (h) x+3)
dx =?
2 dx =?,
x+3)2
R
R
2
2
(i) cos(2x)dx =?,
cos(2x)dx =?
(j) sec (x)dx =?,
sec (x)dx =?
R 1
R 1
1
1
dx =?
(l) 1+x2 dx =?,
dx =?
(k) x dx =?,
x
1+x2
R
R
∗
x2
x2
∗
2
3
(m) 2xe dx =?, R 2xe dx =?
(n) 3x cos(x )dx R=?,
3x2 cos(x3 )dx =?
2x
2x
2x
2x
(o)∗ 1+x
dx =?
(p)∗ 1+x
dx =?
2 dx =?,
4 dx =?,
1+x2
1+x4
Solution
4
(a) x3 dx = d x4 ,
R
x3 dx =
R
4
d x4 =
x4
4
5/2
+ C.
R
R
−2
−2
1
−1
dx = d x−2 , x13 dx = d x−2 = 2x
2 + C.
x3
√
√
R
R
1
1
1/2
√ dx = 2dx
, √x dx = 2 d x = 2 x + C.
x
R (x+1)3
3
(x+1)3
(g) (x + 1)2 dx = d (x+1)
,
d
=
+ C.
3
3
3
R
(i) cos(2x)dx = d sin(2x)
, cos(2x)dx = sin(2x)
+ C.
2
R 21
1
(k) x dx = d ln x, x dx = ln x + C.
x2
x2
(c)
(e)
(m) 2xe dx = de ,
(n) 3x2 cos(x3 )dx = d sin(x3 ),
2x
2
(o) 1+x
2 dx = d ln(1 + x ),
2
dx
2x
−1 2
(x ),
(p) 1+x
4 dx = 1+(x2 )2 = d tan
5.3
R
R
R
(a) Evaluate the following indefinite integrals using the substitution method:
√
Z
Z
Z
Z
√
cos( x)
4
√
sin(3x) dx,
dx,
x3 x4 + 1 dx,
dx.
x
1 + 2x
(b) Calculate the following definite integrals using the substitution rule:
5
0
√
( 3 + 2x) dx,
Z
π/4
0
sin(4t) dt,
Solution
2
5/2
dx
1
−1
−1
(h) (x+3)
= x+3
+ C.
2 dx = d( x+3 ),
(x+3)2
R
(j) sec2 (x)dx = d tan(x),R sec2 (x)dx = tan(x) + C
1
1
−1
−1
(l) 1+x
x, 1+x
x + C.
2 dx = d tan
2 dx = tan
R
R
x2
x2
x2
2xe dx = de R= e + C.
R
3x2 cos(x3 )dx = d sin(x3 ) + sin(x3 ) + C.
R 2x
R
d ln(1 + x2 ) = ln(1 + x2 ) + C.
2 dx =
1+x
R 2x
R
dx = d tan−1 (x2 ) = tan−1 (x2 ) + C.
1+x4
Substitution 1
Z
5/2
(b) x3/2 dx = d 2x5 , x3/2 dx = d 2x5 = 2x5 + C
R
√
R√
3/2
3/2
3/2
xdx = d 2x3 = 2x3 + C.
(d) xdx = d 2x3 ,
R
R
−2x
−2x
−2x
(f) e−2x dx = d e−2 , e−2x dx = d e−2 = e−2 + C.
Z
0
√
π
x cos(x2 ) dx.
(a) (i) Set u = 3x. Then du = 3 dx, and
(ii) Set u =
√
Z
1
sin(3x) dx =
3
Z
sin u du =
− cos u
− cos 3x
=
+ constant.
3
3
x, then du = (1/2)x−1/2 dx, and
√
Z
Z
√
cos x
√
dx = 2 cos u du = 2 sin u = 2 sin x + constant.
x
(iii) Set u = x4 + 1, then du = 4x3 dx and
Z
√
1 Z 1/2
1 u3/2
(x4 + 1)3/2
4
u du =
x x + 1 dx =
=
+ constant.
4
4 3/2
6
3
(iv) Set u = 1 + 2x, then du = 2 dx and
Z
4
dx =
1 + 2x
Z
2
du = 2 ln |u| = 2 ln |1 + 2x| + constant.
u
(b) (i) Set u = 3 + 2x, then du = 2 dx, and the limits of integration change from 0 < x < 5 to
3 < u < 13. (u(0) = 3 + 0 = 3; u(5) = 3 + 10 = 13)
Z
5
0
Z
√
1 2u3/2
1 13 1/2
u du =
3 + 2x dx =
2 3
2
3
"
#
13
3
u3/2 13
=
= 13.89.
3 3
"
#
(ii) Set u = 4t, then du = 4 dt, and the limits of integration change from 0 < t < π/4 to
0 < u < π.
π
Z π/4
Z
1 π
1
1
sin(4t) dt =
sin u du = [− cos u] = .
4 0
4
2
o
0
√
(iii) Set u = x2 , du = 2x dx. Then the limits of integration change from 0 < x < π to
0 < u < π.
π
Z √π
Z
1 π
1
2
x cos(x ) dx =
cos u du = [sin u] = 0.
2 0
2
0
0
5.4
Substitution 2
Compute the following integrals using substitution.
3
1
dy
(a)
1 1 + y2
Z
1
dx
(d)
x ln(x)
p
Z
(g)
Z
(j)
Z
0
5
cos(x) sin (x) dx
sec2 (x)
q
2 + tan(x)
2
3
6
x (x + 1) dx
1
dy
1−y
Z √
(h)
3x + 1 dx
(e)
x2
dx
1 − x3
1
Z
(m)
(b)
Z
dx
Z
3
(k)
dx
4 + 5x
Z
2
dx
(n)
4 + x2
Z
√
ex 1 + 2ex dx
(c)
Z
(f)
Z
(i)
Z
√
(l)
Z
cot(θ) dθ
S
1
k1
dn, k1 , k2 > 0
k2 − n
x
dx
4 + x2
Solution
p
(a)
Z
(b)
Z
Z
1
1
dy = arctan(y)|p1 = arctan(p) − arctan(1).
1 + y2
x2 (x3 + 1)6 dx
Let u = x3 + 1,
du = 3x2 dx. Then I =
Z
x2 (x3 + 1)6 dx =
u6 (du/3) = (1/3)[u7 /7] = (1/21)(x3 + 1)7 + C
Z
Z
Z
√
√
(c)
ex 1 + 2ex dx
Let u = 1 + 2ex , du = 2ex dx. I =
ex 1 + 2ex dx = u1/2 (du/2) =
(1/2)
Z
u1/2 du = (1/2)u3/2 /(3/2) = (1/3)(1 + 2ex )3/2 + C
1
1
dx
u = ln x, du = (1/x) dx. I =
dx = (1/u)du = ln |u| =
x ln(x)
x ln(x)
ln | ln |x|| + C
Z
Z
Z
1
1
1
dy Substitute u = 1 − y, hence du = −dy, and
dy = −
du = − ln |u| =
(e)
1−y
1−y
u
− ln |1 − y| + C.
S
Z S
k1
k2 − 1
(f)
dn = −k1 ln |k2 − n| = −k1 [ln |k2 − S| − ln |k2 − 1|] = k1 ln |
|.
k2 − S
1 k2 − n
1
Z 1
Z
Z
x2
3
2
(g)
dx
u
=
1−x
,
du
=
−3x
dx.
Then
I
=
(1/u)(du/−3)
=
(−1/3)
(1/u) du =
0 1 − x3
(d)
Z
Z
Z
1
−(1/3) ln |u| = −(1/3) ln |1 − x3 | . This integral does not exist since the integrand blows up at
x = 1.
(h)
Z
0
√
3x + 1 dx
u = 3x + 1, du = 3dx. Then I =
(2/9)(3x + 1)3/2 + C
x
√
dx
(i)
4 + x2
√
4 + x2 + C.
Z
Z
u1/2 (du/3) =
1 2
· · u3/2 =
3 3
1 Z −1/2
1 u1/2
u = 4 + x , du = 2x dx. Then I =
= u1/2 =
u
du =
2
2 (1/2)
2
4
(j)
Z
(k)
Z
(l)
Z
cos(x) sin5 (x) dx
3
dx
4 + 5x
cot(θ) dθ =
u = sin x, du = cos x dx. Then I =
u = 4 + 5x, du = 5dx. I = 3
Z
cos(θ)
dθ
sin(θ)
Z
Z
(1/u)(du/5) =
u5 du = u6 /6 = sin6 x/6 + C
3
3
ln |u| = ln |4 + 5x| + C
5
5
Substitute u = sin(θ), hence du = cos(θ) dθ. Then
Z
cot(θ)dθ =
du
= ln |u| + C = ln | sin(θ)| + C.
u
Z
Z
Z
sec2 (x)
1/2
2
q
(m)
dx
u = 2 + tan(x), du = sec x dx. So I = (1/u )du = u−1/2 du =
2 + tan(x)
1/2
√
u
= 2 u = 2(2 + tan x)1/2 + C.
1/2
Z
Z
2
2
2
2
dx
Substitute
u
=
x/2.
Then
x
=
4u
,
du
=
dx/2,
and
dx =
(n)
2
4+x
4 + x2
Z
Z
4
du
du =
= arctan(u) + C = arctan(x/2) + C.
2
4(1 + u )
1 + u2
Z
5.5
Trigonometric Substitution 1
The integral
R
sin(x) cos(x)dx can be done in several ways:
(a) using the substitution u = sin(x) (or u = cos(x)) OR
(b) by first using the trigonometric identity sin(2x) = 2 sin(x) cos(x) and then integrating. Show
that the two answers are equivalent. (Hint: you will find that this is a good opportunity to
review trigonometric identities.)
Solution
(a) u = sin(x), du = cos(x)dx, so I =
Z
sin x cos x dx =
(b) sin(2x) = 2 sin x cos x (trig identity). So I =
u = 2x. Then I = (1/2)
Z
Z
Z
udu = u2 /2 =
1 2
sin x + C.
2
sin(x) cos(x)dx = (1/2)
Z
sin(2x)dx. Let
sin(u)(du/2) = (−1/4) cos(2x) + C 0 . Using the identities cos(2x) =
cos2 x − sin2 x and cos2 x = 1 − sin2 x, we can show that the two results are equivalent: I =
(−1/4)(1 − 2 sin2 x) + C 0 = sin2 x/2 + (C 0 − 1/4). Let C = (C 0 − 1/4) to get the result.
5.6
Trigonometric substitution 2
R √
The integral 01
1 − x2 dx can be done by making the substitution x = sin(u) and dx =
cos(u)du. This is called a trigonometric substitution.
(a) Show that this reduces the integral to
R
cos2 (u) du.
5
(b) Now use the identity cos2 (u) =
integrate.
1 + cos(2u)
to re-express the integral in a simpler form. Then
2
(c) Explain why the answer is the same as the area of 1/4 of a circle of radius 1.
Solution
(a) Let x = sin(u), dx = cos(u)du. Then I =
so that I =
Z
cos u cos u du =
(b) Using the identity we get
Z
cos2 u du.
Z q
1 − sin2 u(cos u du). But 1 − sin2 u = cos2 u,
Z
I = (1/2) [1 + cos(2u)]du = (1/2)[u + (1/2) sin(2u)] = (1/2)[u + sin(u) cos(u)].
√
√
Now we need to express in terms of x, so we use x = sin u, so also 1 − x2 = 1 − sin2 u = cos u
to rewrite the above as
√
I = (1/2)[sin−1 x + x 1 − x2 ]|10 = (1/2)(sin−1 1 − sin−1 0) = (1/2)(π/2 − 0) = π/4
(c) The function y = f (x) =
π/4.
5.7
√
1 − x2 , 0 < x < 1 forms 1/4 of a circle of radius 1, thus area is
Partial Fractions 1
Practice with Integration: Compute the following integrals. Use factoring and/or completing the
square and partial fractions, or some other technique if necessary.
Z
(a)
(c)
Z
1
dx
x2 − x − 20
(b)
−1
dx
2
x + 4x + 14
(d)
Z
Z
3
dx
x2 + 6x + 9
x2
2
dx
− 6x + 8
Solution
dx
dx
=
. Use partial fractions to get
(a) I =
2
x − x − 20 (x − 5)(x − 4) Z 1
1 x − 5 1
1
I=
dx = ln + C.
−
9
x−5 x+4
9
x + 4
Z
Z
3
3
(b)
dx =
dx = −3(x + 3)−1 + C.
2
x + 6x + 9
(x + 3)2
Z
Z
6
Z
√
−1
−1
dx
=
dx.
Now
make
the
substitution
x
+
2
=
(c) I =
10 tan θ
x2 + 4x√+ 14
(x +√
2)2 + 10
Z
sec2 θ
− 10 Z
−1 Z sec2 θ
θ
− 10 sec2 θ
√
√
+C =
dθ
=
dθ
=
dθ
=
−
to get I =
2
2
10 tan! θ + 10
10
tan θ + 1
10 sec2 θ
10
1
x+2
− √ tan−1 √
+ C.
10
10
Z
Z Z
x − 4
1
2
1
2
+ C.
dx = ln dx =
=
−
(d)
x2 − 6x + 8
(x − 4)(x − 2)
x−4 x−2
x − 2
Z
5.8
Partial Fraction 2
The integrals shown below may look very similar, but in fact they lead to quite different results:
(a)
Z
dx
2
a + x2
(b)
Z
a2
dx
− x2
Show that (a) can be reduced to an inverse tangent type integral by a bit of algebraic rearrangement and a substitution of the form u = x/a. Show that (b) can be integrated by factoring the
expression a2 − x2 and using the method of partial fractions.
Solution
Z
(a) I =
Z
dx
=
2
a + x2
adu
= (1/a)
2
a [1 + u2 ]
Z
Z
a2 [1
dx
. Substitute u = x/a, then du = dx/a so I =
+ (x2 /a2 )]
du
= (1/a) arctan(u) + C = (1/a) arctan(x/a) + C.
[1 + u2 ]
dx
1
dx
=
.
Using
partial
fractions,
we
get:
=
a2 − x2
(a + x)(a − x)
(a + x)(a − x)
Z
(1/2a)
1/2a
1
1
1
+
so that I =
+
dx = −(1/2a) ln |a − x| + (1/2a) ln |a +
(a + x) (a − x)
2a (a + x) (a − x)
a+x
| + C.
x| + C = (1/2a) ln |
a−x
(b) I =
5.9
Z
Z
Integration by Parts 1
Use integration by parts to show that
Z
xn ex dx = xn ex − n
Z
xn−1 ex dx
Solution
With u = xn and dv = ex dx, we get du = nxn−1 dx and v = ex , giving
R
n xn−1 ex dx.
7
R
xn ex dx = xn ex −
5.10
Integration by Parts 2
First make a substitution and then use integration by parts to evaluate the following integrals:
(a)
Z
√
sin( x) dx
(b)
Z
4 √
x
e
1
Z
dx
√
π
(c) √
3
π/2
2
x cos(x ) dx
(d)
Z
2
x5 ex dx
Solution
Z
Z
√
√
1
(a) u = x and du = √ dx so sin( x)dx = 2u sin(u)du. Integration by parts: f (u) = 2u
2 x
Z
and g 0 (u) = sin(u). Therefore f 0 (u) = 2 and g(u) = − cos(u) and 2u sin(u)du = −2u cos(u) −
Z
√
√
√
−2 cos(u)du = −2u cos(u) + 2 sin(u) + C = −2 x cos( x) + 2 sin( x) + C.
(b) u =
4 √
x
Z
Z1
e
2
dx =
(c)√ u = x
Z
Z
2
1
2ueu du. With f (u) = 2u and g 0 (u) = eu , we get f 0 (u) = 2 and g(u) = eu , and
2ueu du = 2ueu |21 −
1
π
√
π/2
√
√
1
√ dx. The limits of integration change to 1 and 4 = 1 and 2.
2 x
√
x and du =
2
Z
2
1
2eu du = 4e2 − 2e − 2eu |21 = 2e2 .
→ du = 2x dx.
x3 cos(x2 )dx =
Z
π
π/2
(u/2) cos(u)du. With f (u) = u/2 and g 0 (u) = cos(u), f 0 (u) = 1/2
and g(u) = sin(u), and we get
−π 1
cos(u) π
|π/2 =
− .
2
4
2
2
(d) u = x , du = 2x dx:
u
g(u) = e , and we get
Z
The limits of integration change to π/2 and π, giving us:
Z
Z
5 x2
x e
π
π/2
(u/2) cos(u)du =
= (1/2)
2 u
2 u
Z
(u/2) sin(u)|ππ/2
Z
ueu du. For
0
u
u
Z
x5 ex = (1/2)
R
π/2
−π
sin(u)
du =
+
2
4
ueu du we again use integration by
parts with f (u) = u, f (u) = 1, g (u) = e , and g(u) = e to get:
ueu − eu + C. Altogether, we get
π
u2 eu du. With f (u) = u2 and g 0 (u) = eu , f 0 (u) = 2u,
u e du = u e − 2
0
−
Z
2
2
Z
Compute the following integrals.
8
u
u
ue du = ue −
Z
eu du =
u2 eu du = (1/2)(u2 eu − 2(ueu − eu )) + D =
(1/2)eu (u2 − 2u + 2) + D = (1/2)ex (x4 − 2x2 + 2) + D.
5.11
Z
3
+ 4x + 6
Z
2
dx
(b)
(x + 1)(x + 2)
(a)
(e)
Z
Z
x2
T
0
te−2t dt
(c)
Z
(f)
Z
1
dx
2
x + 6x + 8
π
0
x
x sin( ) dx
2
(d)
Z
(g)
Z
p
1
1
dy
1 − y2
x2 ln(x) dx
Solution
(a)
Z
Z
3
dx =
x2 + 4x + 6
Z
3
dx. With the substitution u = x + 2, this becomes
(x + 2)2 + 2
3
du. We solve this indefinite integral using algebra and substitution:
u2 + 2
3
3Z
1
3Z
du
√
du =
du =
2
2
u +2
2 (u /2) + 1
2 1 + (u/ 2)2
√ √
We now substitute a = u/ 2, 2 da = du, giving us:
√ Z
!
√
da
3
3
3
x+2
3 2
= √ arctan(a) = √ arctan(u/ 2) = √ arctan √
2
1 + a2
2
2
2
2
Z
2
2
A
B
dx
Partial fractions:
=
+
provided
(x + 1)(x + 2)
(x + 1)(x + 2)
(x + 1)
(x + 2)
A(x + 2) + B(x + 1) = 2, for any
x = −1 get A = 2.
! x. Plug in x = −2, get B = −2 plug in
Z
x + 1
2
2
+ C.
dx = 2 ln |x + 1| − 2 ln |x + 2| + C = 2 ln −
Thus I =
(x + 1) (x + 2)
x + 2
(b)
Z
1
dx
Factor denominator: x2 + 6x + 8 = (x + 4)(x + 2). Thus I =
+
6x
+
8
Z
A
B
+
dx. Use Partial fractions. A(x + 2) + B(x + 4) = 1. Plug in x = −2, −4 to
(x + 4) (x + 2)
Z
(−1/2)
(1/2)
find that B = 1/2, A = −(1/2). Integrate I =
+
dx to get I = (−1/2) ln |x +
(x + 4) (x + 2)
1 x + 2 + C.
4| + (1/2) ln |x + 2| + C = ln 2
x + 4
!
1
1
1
1
= (1/2)
+
(d) Note that
=
by partial fractions. There1 − y2
(1 − y)(1 + y)
1−y 1+y
Z
Z
Z
1
1
1
fore,
dy
=
(1/2)(
dy
+
dy) = (1/2)(− ln |1 − y| + ln |1 + y| + C) =
2
1
−
y
1
−
y
1
+
y
1 + y + C 0 . Using this, we can see that the definite integral considered does not ex(1/2) ln 1 − y
Z p
1
dy existed, then this integral would be equal to
ist. First note that, if the integral
1 1 − y2
(c)
Z
x2
9
Z p
1
1
dy (r, p > 1). But using the result obtained, we see that
dy =
the limit lim
r→1 r 1 − y 2
r 1 − y2
!
1 + r 1 + p
− ln = (1/2)(ln |1 + p| − ln |1 − p| − ln |1 + r| + ln |1 − r|). Since the
(1/2) ln 1−p
1 − r
Z p
1
limit limr→1 ln |1 − r| does not exist, the definite integral
dy doesn’t either.
1 1 − y2
Z
(e)
Z
T
0
Z
p
te−2t dt
T
0
te
−2t
Integration by parts: u = t, dv = e−2t dt, hence du = dt and v =
π
Z
π
0
Then
T −1
−t −2t T
dt =
e −
e−2t dt
2
2
0
0
−T −2T
1 −2T T
=
e
−
e
2
4
0
−T −2T 1 −2T 1
=
e
− e
+ .
2
4
4
x
x sin( ) dx
2
0
v = −2 cos( x2 ). Then
(f)
Z
−1 −2t
e .
2
x
x sin
2
Z
Integration by parts: u = x, dv = sin( x2 ) dx, hence du = dx and
dx =
=
π
Z π
−2 cos(x/2)dx
−2x cos(x/2) −
0
0π
0 + 4 sin(x/2) = 4.
0
(g)
Z
Z
x2 ln |x| dx
(Integration by parts: u = ln |x|, du = (1/x) dx and dv = x2 dx, v = x3 /3)
x2 ln |x|dx = ln |x|x3 /3 −
Z
(1/x)(x3 /3)dx = ln |x|x3 /3 −
Z
(x2 /3) dx = ln |x|x3 /3 − x3 /9 + C.
5.12
Integration drills: (Note: some simplifications in a few of these will make your work much easier.)
10
(1)
Z
(4)
Z
(7)
Z
2
(2)
dx
2 + 3x
Z 1
x
dx
(5)
1 + x2
0
Z
1
(8) √
dx
25 − 4x2
1
ln(x) dx
x
2
2t e
1
3t2
Z
dt
2
dx
3 + 4x2
(10)
Z
cos(3x) (1 − sin(3x)) dx
(12)
Z
sec2 (x)
(14)
Z
(16)
Z
(18)
Z
(20)
Z
(22)
Z
(24)
Z
(11)
Z
q
(13)
Z
2x(1 − sin(2x)) dx
(15)
x sin(x + 1) dx
(17)
tan(x) + 1 dx
Z
Z
sec (t) dt
1
dx
(x − 5)(x + 1)
1
dx
(x + 2)(x + 3)
(29)
Z
x sec2 (x) dx (27)
−1
tan (x) dx
Solution
Z
1
ln(x) dx
(1)
x
ln(x)2 /2 + C.
(30)
Z
√
1
dx
2 − 3x2
Z
1
(9) x √
dx
9 + x2
(6)
cos(2x)(1 − sin(2x)) dx
π/2
0
(1 − sin2 (t)) sin(t) dt
sec(x) tan(x) dx (Hint : use your trig relations)
(x sin2 (x) + x cos2 (x))5 dx (Hint : look caref ully!)
(19)
Z
(23)
Z
2x2
3x2 + 1
dx
x2 (x2 + 1)2
1
dx (Hint : try f actoring)
x2 − 3x + 2
Z
x2 sin(3x3 + 1) dx
1
dx (Some algebra, please)
+ 12x + 18
Z
1
(21)
dx
(x + 3)(2x − 1)
2
(26)
(3)
Z
Z
Z
2
xex dx
(25)
(28)
Z
Z
3
dx
x2 + 3x + 15
x2 ex dx
x cos(x) ex dx
With u = ln(x) and du = (1/x) dx,
Z
Z
1
ln(x)dx = udu = u2 /2 + C =
x
(2) Let u = 2 + 3x, du = 3 dx,
2
2
dx =
2 + 3x
3
Z
Z
1
2
2
du = ln |u| + C = ln |2 + 3x| + C.
u
3
3
(3) Let u = 3x3 + 1, du = 9x2 dx.
Z
1
x sin(3x + 1) dx =
9
2
3
Z
1
1
sin(u) du = − cos(u) + C = − cos(3x3 + 1) + C.
9
9
11
(4) Let u = 3t2 , du = 6t dt. Then the new limits of integration are 3(1)2 and 3(2)2 , or 3 and 12.
Z
2
2t e
1
3t2
1
dt =
3
Z
12
3
1 u=12 1 12
= (e − e3 ).
e du = eu 3 u=3
3
u
(5) Let u = 1 + x2 , du = 2x dx. The limits of integration change to 1 + 02 = 1 and 1 + 12 = 2.
Z
1
0
x
1
dx =
2
1+x
2
(6)
I=
Let u =
q
3
x
2
→ du =
q
3
2
Z
√
dx:
Z
2
1
u=2
1
1
1
1
du = ln u
= (ln 2 − ln 1) = ln 2.
u
2
2
2
u=1
1 Z
1
1
q
√
dx
=
dx.
2
2 − 3x
2
1 − (3/2)x2

s
√ Z
√ !
1
1 2
3
6
1
1
√
x + C = √ arcsin
x +C
I=√ √
du = √ arcsin 
2
2
2
2 3
1−u
3
3
(7)
Z
Let tan u =
√2
3
1
√
3
Z
2
2
dx =
2
3 + 4x
3
x, then sec2 u · du =
√2
3
sec2 u
1
du = √
2
1 + tan u
3
Z
Z
1
dx
1 + 43 x2
dx ⇒ dx =
√
2 3
3 2
· sec2 u du
!
1
2x
1
du = √ u + C = √ tan−1 √ + C.
3
3
3
(8) We make the substitution 2x = 5u, 2dx = 5du:
Z
Z
1
1
2x
1
5/2
√
+ C.
dx = √
du = sin−1 u + C = sin−1
2
2
2
2
5
25 − 4x
25 − 25u
(9) We make the substitution u = 9 + x2 , du = 2xdx:
Z
√
x
dx =
9 + x2
Z
√
1
√ du = u1/2 + C = 9 + x2 + C.
2 u
(10) We use the trig identities sin 2θ = 2 sin θ cos θ and cos(2θ) = 1 − 2 sin2 (θ):
Z
cos(3x) (1 − sin(3x)) dx =
=
Z
Z
(cos(3x) − cos(3x) sin(3x)) dx
(cos(3x) dx −
12
Z
sin(6x)
) dx
2
sin(3x) cos(6x)
+
+C
3
12
4 sin(3x) + 1 − 2 sin2 (3x)
+C
=
12
−1
=
(sin2 (3x) − 2 sin(3x)) + C 0
6
=
(11) We use the trig identity sin 2θ = 2 sin θ cos θ:
Z
Z
cos(2x)(1 − sin(2x)) dx =
cos 2x dx −
Z
cos 2x sin 2x dx
sin 2x
sin 4x
−
dx
2
2
sin 2x cos 4x
=
+
+ C.
2
8
Z
=
(12) Let u = tan x + 1, du = sec 2 x dx.
Z
sec2 (x)
q
tan(x) + 1 dx =
Z
√
u du =
2
2 3/2
u + C = tan(x + 1)3/2 + C.
3
3
(13) Let u = cos t, du = − sin t dt. Set new limits of integration with respect to u: t = 0, u = 1.
t = π/2, u = 0.
Z
π/2
0
2
(1 − sin (t)) sin(t) dt =
Z
π/2
0
2
cos t sin t dt = −
Z
0
1
2
u du =
(14)
Z
2x(1 − sin(2x)) dx =
Z
2x dx −
= x2 −
Z
Z
Z
1
0
u3 u=1 1
u du = = .
3 u=0 3
2
2x sin 2x dx
2x sin 2x dx
This is now done by parts, with u = 2x and dv = sin(2x) dx. Therefore du = 2 dx and
v = (−1/2) cos(2x).
x2 −
Z
2x sin 2x dx = x2 − −x cos(2x) −
= x2 + x cos 2x −
Z
− cos(2x) dx
sin 2x
+ C.
2
(15) We make the substitution u = cos x, du = − sin x dx:
Z
sec(x) tan(x) dx =
Z
Z
1
−du
1
sin x
dx
=
+ C = sec x + C.
= +C =
2
2
cos x
u
u
cos x
13
(16) This is done by parts, with u = x and dv = sin(x + 1) dx. Therefore du = dx and
v = − cos(x + 1):
Z
x sin(x + 1) dx = −x cos(x + 1) +
Z
cos(x + 1) dx = −x cos(x + 1) + sin(x + 1) + C.
(17) Note that sin2 x + cos2 x = 1.
Z
(x sin2 (x) + x cos2 (x))5 dx =
Z h x sin2 (x) + cos2 (x)
i5
dx =
Z
x5 dx =
x6
+ C.
6
(18)
Z
(19)
Z
sec2 (t) dt = tan(t) + C
1
dx =
2
2x + 12x + 18
Z
1
1
dx = − (x + 3)−1 + C.
2
2(x + 3)
2
(20) Partial fractions:
Z
1Z
1
dx =
(x − 5)(x + 1)
6
1
1
−
x−5 x+1
(21) Partial fractions:
Z
1
1Z
dx = −
(x + 3)(2x − 1)
7
(22) Partial fractions:
Z
(23)
Z
1 x − 5 dx = ln + C.
6
x + 1
2
1
1 x + 3 −
dx = − ln + C.
x + 3 2x − 1
7
2x − 1 1
1
1
=
−
.
(x + 2)(x + 3)
x+2 x+3
1
dx =
(x + 2)(x + 3)
3x2 + 1
dx =
x2 (x2 + 1)2
Z
Z
1
dx −
x+2
Z
1
dx
x+3
x + 2
+ C.
= ln |x + 2| − ln |x + 3| + C = ln 3x2 + 1
dx =
[x(x2 + 1)]2
Z
3x2 + 1
dx.
[x3 + x]2
x+3
Let u = x3 + x, du = (3x2 + 1) dx,
Z
Z
1
1
1
3x2 + 1
+ C.
dx =
du = − + C = − 3
3
2
2
[x + x]
u
u
x +x
14
(24) Partial fractions:
Z
(25)
Z
Let u =
4
17
Z
√2
51
1
dx =
2
x − 3x + 2
3
dx =
2
x + 3x + 15
x+
3
2
, du =
√2
51
Z
Z x − 2
1
1
+ C.
= ln −
x−2 x−1
x − 1
3
x+
3
2
dx,
2
+
51
4
dx =
3
51
4
Z
1
4
51
x+
3
2
2
dx.
+1
√ Z
4 51
1
dx =
du
2
2
3
4
17 2
u +1
+
1
x
+
51
2
√
2 51
tan−1 (u) + C
=
17
√
!
2 51
3
2
−1
√
=
x+
+ C.
tan
17
2
51
1
(26) By parts, with u = x, du = dx, dv = sec2 x dx, and v = tan x:
Z
x sec2 (x) dx = x tan(x) −
Z
tan(x) dx = x tan(x) −
Z
sin(x)
dx.
cos(x)
Use substitution, with u = cos(x) and du = − sin(x) dx. Then the integral becomes
x tan(x) +
Z
du
= x tan(x) + ln | cos(x)| + C.
u
(27) We make the substitution u = x2 , du = 2x dx:
Z
xe
x2
dx =
Z
1 u
1
1 2
e du = eu + C = ex + C.
2
2
2
(28) This is done by using parts twice. The first time, u = x2 and dv = ex dx, so that du = 2x dx
and v = ex :
Z
Z
2 x
2 x
x e dx = x e − 2xex dx
The second time, u = 2x, du = 2 dx, dv = ex dx, and v = ex .
2 x
x e −
Z
x
2 x
x
2xe dx = x e − 2xe −
Z
x
2e dx
= x2 ex − 2xex + 2ex + C.
15
(29) This is done by parts, with u = tan−1 (x) and dv = dx. Then du =
Z
−1
−1
tan (x) dx = x tan (x) −
Z
1
1+x2
dx, and v = x.
x
dx
1 + x2
Now use substitution, letting u = 1 + x2 . Then du = 2x dx.
x tan−1 (x) −
Z
x
1 Z du
1
−1
−1
dx
=
x
tan
(x)
−
=
x
tan
(x)
−
ln(1 + x2 ) + C.
2
1+x
2
u
2
(30) x cos(x) ex dx
Let u = x cos x, dv = ex dx, du = (cos x − x sin x) dx, v = ex
R
I=
Z
x
x
x cos x e dx = x cos x e −
Z
x
e cos x dx +
Z
x sin x ex dx.
Let u = x sin x, dv = ex dx, du = (sin x + x cos x) dx, v = ex ,
x
I = x cos x e −
Z
x
x
e cos x dx + x sin x e −
⇒ 2I = x cos x ex −
Z
Z
x
e sin x dx −
ex cos x dx + x sin x ex −
Let u = sin x, dv = ex dx, du = cos x dx, v = ex ,
x
⇒ 2I = x cos x e −
⇒I =
Z
x
x
x
Z
Z
x cos x ex dx ⇐ I.
ex sin x dx.
e cos x dx + x sin x e − e sin x −
Z
1
(x cos x ex + x sin x ex − ex sin x) + C.
2
16
x
e cos x dx .