Calculus 2
Lia Vas
Integration by Parts
Using integration by parts one transforms an integral of a product of two functions into a simpler
integral. Divide the initial function into two parts called u and dv (keep dx in dv part). Then apply
the following rule.
Integration by parts:
Z
udv = uv −
You were successful in choosing u and dv initially if the resulting integral
the initial integral. If it is not, go back and rethink your choice of u and dv.
R
Z
vdu
vdu is simpler that
Below are some hints on how to decompose the initial integral into udv:
a) Integrals with the product of a polynomial and eax . Try u = polynomial.
b) Integrals with polynomial and sin ax (or cos ax). Try u = polynomial.
c)
R ax
e sin bx
dx or eax cos bx dx. You can start with u = eax . You will need to do integration
by parts twice and will end up with your initial integral after the second time. Solve for your
initial integral then.
R
d) Integrals with logarithmic functions. Try u = logarithmic function.
e) Integrals with inverse trigonometric functions. Try u = inverse trigonometric function.
The formula udv = uv − vdu is really the product rule in disguise. To see why the integration
by parts formula is true, start with the product rule (uv)0 = u0 v + v 0 u that can beR also written
du
dv
d
d
(uv) =
v + dx
u. Integrating
the product
ruleR with respect to x, we have
that dx
(uv)dx =
as
R dudx
R dvdx
R
R
R
R
vdx
+
udx
⇒
uv
=
duv
+
dvu
=
vdu
+
udv.
Solve
for
the
term
udv
on
the
right side
dx
dx
R
R
and obtain that uv − vdu = udv.
R
R
For definite integrals, the formula becomes
Z b
a
udv =
uv|ba
Practice Problems. Evaluate the integrals:
−
Z b
a
vdu.
1.
2.
3.
Z
xex dx
Z
xe2x dx
Z
x2 ex dx
Hypothesize on the number of integration by parts you would need to evaluate
R n
x sin xdx.
4.
Z 1
R
xn ex dx or
xe−x dx
0
5.
6.
Z
Z
2x sin 3x dx
(2x + 5) sin(2x + 5) dx
7.
Z
8.
ex sin x dx
Z
9.
Z
10.
12.
Z
Z
ln x
dx
x2
ln(2x − 1)
dx
(2x − 1)2
Z
11.
ln x dx
tan−1 x dx
3 sin−1 2x dx
Solutions:
1. Following the first hint, start by u = x and dv = ex dx. Then
du = dxRand v = ex dx
= ex .
R
R
Then Ruse the formula for integration by parts and obtain xex dx = u dv = uv − v du =
xex − ex dx. Note that this last integral is simpler that the initial integral indicating that you
are on the right path. The last integral is equal to ex so your final answer is xex − ex + c.
R
2. Following the first hint, start by u = x and dv = e2x dx. Then du = dx and v = eR2x dx. To
get
v, you can use the substitution w = 2x ⇒ dw = 2dx ⇒ dw
= dx and so v = e2x dx =
2
1 w
1 2x
1R w
e dw = 2 e = 2 e .
2
R
Then xe2x dx = u dv = uv − v du = 12 xe2x − 12 e2x dx. Note that this last integral is
simpler that the initial integral indicating that you are on the right path. In fact, the last
integral is the same as the one you evaluated
when finding v so it is equal to 12 e2x . Thus, the
R
initial integral is equal to 12 xe2x − 21 e2x dx = 12 xe2x − 12 21 e2x + c = 12 xe2x − 14 e2x + c
R
R
R
R
3. RStart by u R= x2 and dv R= ex dx. Then du
= 2xdx and v =R ex dx = ex . Then you have
R x
2 x
2 x
x e dx = u dv = uv − v du = x e − e 2x dx = x2 ex + 2xex dx. For this last integral,
you need to use integration by parts again. Take u = 2xR and dv = ex dx (alternatively,
R
factor R2 out and take u = x). Then du = 2dx and v =R ex dx = ex and so 2xex dx =
2xex − 2ex dx = 2xex −2ex . This gives you the final answer x2 ex dx = x2 ex −(2xex −2ex )+c =
x2 ex − 2xex + 2ex + c.
R
4. Start by u = x and dv = e−x dx. Then du = dx and
v = e−x dx. ToRget v, you can use the
R
−x
−x
substitution w = −x and obtain v = −e . Then xe dx = −xe−x + e−x dx = −xe−x − e−x .
Substituting the bounds you get (−xe−x − e−x )|10 = −e−1 − e−1 + 1 = 1 − 2e ≈ .264.
R
5. Following
the second hint, you can start by u = 2x and dv = sin 3x dx. Then du = 2dx and
R
= dx and so v =
vR = sin 3x dx.R To find v, use the substitution w = 3x ⇒ dw =R3dx ⇒ dw
3
1
−1
−1
sin 3x dx = 3 sin wdw = 3 cos w = 3 cos 3x. Then you have 2x sin 3x dx = −2
x cos 3x −
3
−2 R
cos 3x dx. This last integral is similar to the one used to obtain v and it is equal to 13 sin 3x.
3
x cos 3x + 32 31 sin 3x = −2
x cos(3x) + 29 sin(3x) + c.
Thus the initial integral is equal to −2
3
3
6. Method (1) Use substitution
w = 2x + 5 first to simplify the integral. In this case, dw =
R 1
dw
2dx ⇒ 2 = dx. Obtain 2 w sin w dw. Using the integration by parts with u = 12 w and
w cos w + 12 sin w + c = −1
(2x + 5) cos(2x + 5) + 12 sin(2x + 5) + c.
dv = sin wdw we obtain −1
2
2
Method (2) Use the
integration by parts with u = 2x + 5 and dv = sin(2x + 5). In this case,
R
du = 2dx and v = sin(2x + 5)dx. To findRv, you can use substitution w = 2x + 5. In this
case, dw = 2dx ⇒ dw
= dx. Obtain v = R12 sin wdw == −1
cos w = −1
cos(2x + 5). So, the
2
2
2
−1
−1
integral becomes 2 (2x + 5) cos(2x + 5) − 2 cos(2x + 5) 2dx. This last integral simplifies to
R
cos(2x + 5)dx (factor the negative out), you can evaluate it similarly as when finding v and
obtain 12 sin(2x + 5). So, the final answer is −1
(2x + 5) cos(2x + 5) + 21 sin(2x + 5) + c.
2
7. Note that the third hit applies to this problem. Let us denote the
initial integral by I. You
R
x
x
can start
by u = e an dv = sin
xdx. Then dy = e and v = sin xdx = − cos x so that
R
R
I = ex sin xdx = −ex cos x + ex cos xdx. Use the integration by parts again for this last
integral. With u = ex Rand dv = cos xdx, you obtain du = ex dx and v = sin x so that
R x
e cos xdx = ex sin x − ex sin xdx. This last integral is our initial integral that we denoted
by I. Thus
I = −ex cos x +
Z
ex cos xdx = −ex cos x + ex sin x −
Z
ex sin x = ex cos x + ex sin x − I.
Note that this gives you the equation I == ex cos x + ex sin x − I. Solving for I gives you
1
2I = ex cos x + ex sin x ⇒ I = (ex cos x + ex sin x)
2
So, the final answer is I = 21 ex cos x + 21 ex sin x + c.
8. Following
the fourth hint,
start by u = ln xR and dv = dx. Then du = x1 dx and v =
R
R 1
Thus ln xdx = x ln x − x xdx = x ln x − dx = x ln x − x + c.
9. Start by
u = ln x and dv
= 1 dx = x−2 dx Rso that du = x1 dx and v =
R −1 1x2
R ln x
x
x
− ln x
+ x−2 dx = − ln
− x1 + c.
Then x2 dx = x − x x dx = − ln
x
x
R
R
dx = x.
x−2 dx = −x−1 =
−1
.
x
10. Method (1) Use substitution
w = 2x−1 first to simplify the integral. In this case, dw = 2dx ⇒
dw
1 R ln w
=
dx.
Obtain
dw.
Then
use integration by parts with u = ln w and dv R= dw
. Then
2
2 R w2
w2
1
−1
1
1 −1
1
−2
−1
du = w dw and v = w dw = −w = − w . So, the integral becomes 2w ln w − 2 w w dw =
R
ln(2x−1)
− ln w
ln w
ln w
1
1
+ 12 w12 dw = −2w
− 12 w−1 = −2w
− 2w
+ c = −2(2x−1)
− 2(2x−1)
+ c.
2w
dx
Method (2) Use integration by parts with u = ln(2x − 1) and dv = (2x−1)
2 . In this case,
R
1
−2
du = 2x−1 2dx and v = (2x − 1) dx You can use substitution w = 2x − 1 to find v. In this
R
−1
case, dw = 2dx ⇒ dw
= dx. Obtain v = 12 w−2 dw = −1
w−1 = −1
= 2(2x−1)
. So, the integral
2
2
2w
ln(2x−1)
2
1
−1
−1
ln(2x − 1) − 2x−1
dx = −2(2x−1)
+ (2x−1)
becomes 2(2x−1)
2 dx. This last integral is
2(2x−1)
−1
the same one you evaluated when finding v, and so it is 2(2x−1) . Thus, the final answer is
R
− ln(2x−1)
2(2x−1)
−
1
2(2x−1)
R
+ c.
1
−1
dx =
11. Following the fourth hint, start by u = tan
x and dv = dx. Thus du = 1+x
2 dx and v =
R
x
−1
2
x. So, the integral becomes x tan x − 1+x2 dx. Use the substitution w = 1 + x for this last
R x
R x dw
integral and obtain that 1+x
= 21 ln |w| = 21 ln(1 + x2 ). So, the initial integral is
2 dx =
w 2x
equal to x tan−1 x − 21 ln(1 + x2 ) + c.
R
12. You can apply two methods to this problem as well: (1) use the substitution w = 2x to
simplify first and then use the integration by parts, or (2) use the integration by parts with
u = sin−1 (2x) right away.
Using method (1), the integral reduces to 32 sin−1 w dw. Then, using the integration by
√
parts with u = sin−1 w, you will obtain 32 w sin−1 w + 32 1 − w2 + c. Thus, the final answer is
√
3x sin−1 2x + 23 1 − 4x2 + c.
R
Using method (2), you have that u = sin−1 (2x) and dv = 3dx so that du = √
1
2dx
1−(2x)2
=
2
and v = 3dx = 3x. Thus the integral is 3x sin−1 2x − 3x √1−4x
2 dx. Evaluate this
R
R
1
dw
−3
2
−1/2
integral using the substitution w = 1−4x and obtain 6x √w −8x = 4 w
dw = −3
2w1/2 =
4
√
√
− 32 1 − 4x2 . So, the final answer is 3x sin−1 2x + 23 1 − 4x2 + c.
√ 2
dx
1−4x2
R
R