Practice Problems: Integration by Parts (Solutions)
Written by Victoria Kala
[email protected]
November 25, 2014
The following are solutions to the Integration by Parts practice problems posted November 9.
R
1. ex sin xdx
Solution: Let u = sin x, dv = ex dx. Then du = cos xdx and v = ex . Then
Z
Z
ex sin xdx = ex sin x − ex cos xdx
Now we need to use integration by parts on the second integral. Let u = cos x, dv = ex dx.
Then du = − sin xdx and v = ex . Then
Z
Z
ex sin xdx = ex sin x − ex cos x − ex sin xdx
The right integral is the same as the one we started with! Move it over:
Z
2 ex sin xdx = ex sin x − ex cos x
And divide by 2:
Z
ex sin xdx =
1 x
(e sin x − ex cos x)
2
This is our final solution, so make sure to add your constant C:
Z
1
ex sin xdx = (ex sin x − ex cos x) + C
2
2.
R
(sin−1 x)2 dx
Solution: Let u = (sin−1 x)2 , dv = dx. Then du =
Z
2√
sin−1 x
dx,
1−x2
(sin−1 x)2 dx = x(sin−1 x)2 −
Z
v = x. Then
2x sin−1 x
√
dx
1 − x2
We need to use a substitution on the last integral. Let w = sin−1 x. Then dw =
and x = sin w. Just looking at the last integral, we have:
Z
Z
2x sin−1 x
√
dx = 2w sin wdw
1 − x2
1
√ 1
dx
1−x2
We can use integration by parts on this last integral by letting u = 2w and dv = sin wdw.
Tabular method makes it rather quick:
Z
2w sin wdw = 2w cos w + 2 sin w
At this point you can plug back in w:
Z
2w sin wdw = 2 sin−1 x cos (sin−1 x) + 2 sin (sin−1 x)
OR you can look at the triangle formed by our substitution for w. Since x =√sin w then the
hypotenuse will be 1, the opposite side will be x and the adjacent side will be 1 − x2 . Then
Z
p
2w sin wdw = 2 1 − x2 sin−1 x + 2x
Either of these solutions is fine. So then our integral will look like either one of the solutions
below:
Z
(sin−1 x)2 dx = x(sin−1 x)2 − (2 sin−1 x cos (sin−1 x) + 2 sin (sin−1 x)) + C
Z
p
(sin−1 x)2 dx = x(sin−1 x)2 − (2 1 − x2 sin−1 x + 2x) + C
3.
R
x tan2 xdx
Solution: Use the identity tan2 x = sec2 x − 1:
Z
Z
Z
Z
2
2
2
x tan xdx = x(sec x − 1)dx = x sec xdx − xdx
The last integral is no problemo. The first integral we need to use integration by parts. Let
u = x, dv = sec2 x. Then du = dx, v = tan x, so:
Z
Z
2
x sec xdx = x tan x − tan xdx
You can rewrite the last integral as
− ln | cos x|, so:
Z
R
sin x
cos x dx
and use the substitution w = cos x.
x sec2 xdx = x tan x + ln | cos x|
Plug that into the original integral:
Z
1
x tan2 xdx = x tan x + ln | cos x| − x2 + C
2
2
R
tan xdx =
4.
R1
0
t cosh tdt
Solution: This is quick with tabular method. Let u = t, dv = cosh t:
Z 1
1
t cosh tdt = t sinh t − cosh t = sinh(1) − cosh(1) + cosh(0)
0
0
You can leave your answer like this. If you want to evaluate it further, remember that
ex − e−x
ex + e−x
sinh x =
and cosh x =
. Then we see that sinh(1) = 12 (e1 − e−1 ), cosh(1) =
2
2
1 1
−1
), and cosh 0 = 1. Then
2 (e + e
Z
1
t cosh tdt = sinh(1) − cosh(1) + cosh(0) = 1 −
0
1
e
5.
R
z 3 ez dx
Solution: Tabular is the way to go with this baby. Let u = z 3 , dv = ez dz. Then
Z
z 3 ez dx = z 3 ez − 3z 2 ez + 6zez − 6ez + C = ez (z 3 − 3z 2 + 6z − 6) + C
6.
R √3
1
arctan(1/x)dx
Solution: Let u = arctan(1/x), dv = dx. Then du =
√
Z
3
−dx
x2 +1
√3 Z
arctan(1/x)dx = x arctan(1/x) +
1
1
(using chain rule), v = x:
√
3
1
x2
x
dx
+1
The last integral you can use the substitution w = x2 + 1. Then:
√
√3
1
2
arctan(1/x)dx = x arctan(1/x) + ln (x + 1)
2
1
1
√
√
√
1
1
3π 1
π
+ ln 2 −
= 3 arctan ( 3) + ln 4 − arctan (1) + ln 2 =
2
2
3
2
4
Z
3
7.
R
cos x ln (sin x)dx
Solution: We first need to do a substitution. Let w = sin x, then dw = cos xdx:
Z
Z
cos x ln (sin x)dx = ln wdw
3
Next use integration by parts with u = ln w, dv = dw. Then du = w1 dw, v = w:
Z
Z
ln wdw = w ln w − dw = w ln w − w
We need to plug back in w:
Z
cos x ln (sin x)dx = sin x ln(sin x) − sin x + C
8.
R 2 (ln x)2
dx
1
x3
You can do this problem a couple different ways. I will show you two solutions.
Solution I: First do the substitution w = ln x. Then dw = x1 dx and x = ew . Then
Z
1
2
(ln x)2
dx =
x3
Z
ln 2
0
w2
dw =
e2w
Z
ln 2
w2 e−2w dw
0
Tabular is easy on this guy:
Z ln 2
e−2w
1 ln 2
w2 −2w w −2w 1 −2w ln 2
2
2 −2w
− e
− e
=−
w +w+
w e
dw = − e
2
2
4
2
2 0
0
0
1
3
=−
(ln 2)2 + ln 2 +
8
2
1
Solution II: Start of with integration by parts. Let u = (ln x)2 , dv = 3 dx. Then du =
x
1
2 ln x
dx, v = − 2 :
x
2x
Z 2
Z 2
(ln x)2
(ln x)2 2
ln x
dx = −
+
dx
3
2 1
x
2x
x3
1
1
1
1
1
Do integration by parts again. Let u = ln x, dv = 3 dx. Then du = dx, v = − 2 :
x
x
2x
Z
Z 2
ln x
ln x 2
ln x
1
1
2
dx
=
−
+
dx
=
−
−
3
x3
2x2 1
2x2
4x2 1
1 2x
Plugging this into the original integral we get:
Z 2
(ln x)2
ln x
1
1
1 2
(ln x)2
2
2
dx = −
− 2 − 2 = − 2 (ln x) + ln x +
x3
2x2
2x
4x
2x
2 1
1
1
1
3
=−
(ln 2)2 + ln 2 +
8
2
4
9.
R
cos
√
xdx
√
√
1
x. Then dw = √ dx ⇒ 2 xdw = dx ⇒ 2wdw =
2 x
Z
Z
√
cos xdx = 2w cos wdw
Solution: First do the substitution w =
dx:
Using tabular with u = 2w, dv = cos wdw we get:
Z
2w cos wdw = 2w sin w + 2 cos w + C
Plug back in w to get the final solution:
Z
√
√
√
√
cos xdx = 2 x sin x + 2 cos x + C
R √π
10. √
π/2
θ3 cos(θ2 )dθ
Note: There was a typo on the original, it should be dθ instead of dx.
R √π
R √π
θ3 cos(θ2 )dθ = √
θ · θ2 cos(θ2 )dθ. Then use the substitution
Solution: Rewrite: √
π/2
π/2
w = θ2 , so we have dw = 2θdθ:
√
Z
√
π
θ3 cos(θ2 )dθ =
π/2
1
2
Z
π
w cos wdw
π/2
Tabular makes this easy with u = w, dv = cos wdw:
Z
π
1 π
1 π
1
=− −
w cos wdw = (w sin w + cos w) 2 π/2
2
2
4
π/2
11.
R
x ln(1 + x)dx
Solution: Use the substitution w = 1 + x. Then dw = dx and x = w − 1:
Z
Z
x ln(1 + x)dx = (w − 1) ln wdw
1
Next use integration by parts with u = ln w, dv = (w − 1)dw. Then du = dw and v =
w
1 2
w −w :
2
Z
Z 1 2
1
(w − 1) ln wdw =
w − w ln w −
w − 1 dw
2
2
5
The right integral is straightforward, so
Z
1 2
1
(w − 1) ln wdw =
w − w ln w − w2 + w + C
2
4
Next, plug back in w:
Z
1
1
x ln(1 + x)dx =
(1 + x)2 − (1 + x) ln (1 + x) − (1 + x)2 + 1 + x + C
2
4
This answer is fine. You can simplify it a bit more for kicks and giggles:
Z
1
1
1
x ln(1 + x)dx = (x2 − 1) ln (1 + x) − x2 + x + C
2
4
2
12.
R
sin(ln x)dx
1
Solution: Use the substitution w = ln x. Then dw = dx ⇒ xdw = dx ⇒ ew dw = dx since
x
x = ew from our substitution. Then we have:
Z
Z
sin(ln x)dx = ew sin wdw
This is the same as Problem #1, so
Z
1
ew sin wdw = (ew sin w − ew cos w) + C
2
Plug back in w:
Z
sin(ln x)dx =
1
(x sin (ln x) − x cos (ln x)) + C
2
13.
R
√
x3 1 + x2 dx
You can do this problem a couple different ways. I will show you two solutions.
Solution I: You can actually do this problem without using integration by parts. Use the
substitution w = 1 + x2 . Then dw = 2xdx and x2 = w − 1:
Z
Z
Z
Z
p
p
√
1
1
x3 1 + x2 dx = x · x2 1 + x2 dx =
(w − 1) wdw =
(w3/2 − w1/2 )dw
2
2
=
1 5/2 1 3/2
1
1
w
− w
+ C = (1 + x2 )5/2 − (1 + x2 )3/2 + C
5
3
5
3
Solution II: You can use integration by parts as well, but it is much more complicated. Rewrite
the integral:
Z
Z
p
p
1 2
x3 1 + x2 dx =
x · 2x 1 + x2 dx
2
6
√
Let u = 12 x2 , dv = 2x 1 + x2 dx. Then du = xdx, v = 23 (1 + x2 )3/2 (using a substitution on
dv):
Z
Z
p
1
2
1 2
x(1 + x2 )3/2 dx
x · 2x 1 + x2 dx = x2 (1 + x2 )3/2 −
2
3
3
You can use a substitution on the last integral:
Z
p
1
2
1 2
x · 2x 1 + x2 dx = x2 (1 + x2 )3/2 − (1 + x2 )5/2 + C
2
3
15
14. Find the area between the given curves: y = x2 ln x, y = 4 ln x
Solution: We need to find when the two curves intersect, so set them equal to each other:
x2 ln x = 4 ln x ⇒ (x2 − 4) ln x = 0 ⇒ (x − 2)(x + 2) ln x = 0
The solutions to this equation are x = −2, 2, 1. But, x = −2 isn’t in our domain (since ln x
has the domain (0, ∞)), so we are going to toss that solution out. This means we are going
to integrate from x = 1 to x = 2. You can just guess which function is on the top or bottom:
Z 2
Z 2
Z 2
2
A=
(top function − bottom function) dx =
(4 ln x − x ln x)dx =
(4 − x2 ) ln xdx
1
1
1
= 4x − 31 x3 :
Using integration by parts, let u = ln x, dv = (4 − x )dx. Then du =
Z 2
2 Z 2 1 2
1 3
2
4 − x dx
(4 − x ) ln xdx = 4x − x ln x −
3
3
1
1
1
1
16
29
1
2
ln 2 −
=
4x − x3 ln x − 4x + x3 =
3
9
3
9
1
1
x dx, v
2
15. Use the method of cylindrical shells to the find the volume generated by rotating the region
bounded by the given curves about the specified axis: y = e−x , y = 0, x = −1, x = 0 about
x = 1.
Solution: Draw a picture of what is happening. Recall that the volume for a cylinder is
V = 2πRH. In this scenario, R = 1 − x and H = e−x (since H is the top function minus the
bottom function). x is going from -1 to 0:
Z 0
V =
2π(1 − x)e−x dx
−1
Using tabular is pretty quick with u = 1 − x, dv = e−x dx:
Z 0
0
2π(1 − x)e−x dx = 2πxe−x = 2πe
−1
−1
7