Integration By Parts
Alvin Lin
Calculus II: August 2016 - December 2016
Integration By Parts
Product Formula:
d
(f g) = f g 0 + f 0 g
dx
Integrate on both sides:
f g 0 dx +
Z
f 0 gdx
f g 0 dx = f g −
Z
f 0 gdx
Z
Z
fg =
Z
Z
Or with uv as variables:
udv = uv −
vdu
For difficult integrals, this method can be used to rewrite the integral into
an friendlier form.
Practice problem 1
Z
x cos(5x)dx
Let : f (x) = x g 0 (x) = cos(5x)dx
f 0 (x) = dx g(x) =
1
sin(5x)
5
Z
Z
f g 0 dx = f g −
Z
f 0 gdx
x sin(5x) 1 Z
x cos(5x)dx =
−
sin(5x)dx
5
5
x sin(5x) 1 cos(5x)
− (−
)+C
=
5
5
5
1
1
= x sin(5x) +
cos(5x) + C
5
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Practice problem 10
Z
√
ln( x)dx
1Z
1Z
ln(x)dx =
1 ln(x)dx
2
2
Let : g 0 (x) = 1dx f (x) = ln(x)
g(x) = x f 0 (x) =
Z
1
dx
x
Z
√
1
1
x ln(x) − x dx
ln( x)dx =
2
x
Z
1
x ln(x) − 1dx
2
x ln(x) − x
+C
=
2
Practice problem 12
Z
Z
tan−1 2udu
tan−1 2udu =
Z
1 tan−1 2udu
Let : g 0 (u) = 1du f (u) = tan−1 2u
g(u) = u f 0 (u) = 2
2
1
du
1 + 4u2
2u
du
1 + 4u2
After using integration by parts, we can solve the new integral with usubstitution.
Z
2u
du
1 + 4u2
Z
−1
tan
−1
2udu = u tan
2u −
Z
Let : x = 1 + 4u2
dx = 8udu
Z
2u dx
x 8u
1 Z dx
4
x
1
ln(x) + C
4
=
Therefore:
1
ln(1 + 4u2 ) + C
4
Z
tan−1 2udu
= u tan−1 2u −
1
ln(1 + 4u2 ) + C
4
Practice problem 32
(ln(x))2
dx
x3
Z 2
1
Let : u = ln(x)
eu = x
The limit of the integral from 1 to 2 changes to e1 to e2
eu du = dx
Z e2
e
=
Z e2
u2 eu du
(eu )3
u2 e−2u du
e
3
Now we can use integration by parts:
Let : f (x) = u2
g 0 (x) = e−2u
e−2u
−2
f 0 (x) = 2udu g(x) =
Z e2
u2 e−2u du = u2
e
Z
e−2u
e−2u
du
− 2u
−2
−2
u2 e−2u Z
+ ue−2u du
−2
In order to solve the integral properly, you need to use integration by parts
again:
Let : f (x) = u g 0 (x) = e−2u du
=
e−2u
f (x) = 1du g(x) =
−2
0
u2 e−2u
e−2u
u2 e−2u Z
+
− (1)
du
−2
−2
−2
u2 e−2u 1 Z −2u
u2 e−2u
+
+
e du
−2
−2
2
Z
1
−2u
2 −2u
e du
−u e
+
2
e2
1 e−2u
− u2 e−2u + (
)
2 −2
2
−e4 e−2e −
−2e2
e
4
e
− − e2 e−2e − (
e−2e
)
4
2
2
−e4 e−2e −
= −e
−8e2
e−2e
e−2e
+ e2 e−2e +
4
4
+e
−4e
4
+
e−2e
2 +e−2e
4
Practice problem 38
Z
cos(ln(x))dx
Let : u = ln(x)
x = eu
dx = eu du
Z
u
cos(u)e du =
Z
eu cos(u)du
We can describe this integral in a more general form as:
I=
Z
eax cos(bx)dx
where a and b are both 1. Using integration by parts where:
f (x) = eax
g 0 (x) = cos(bx)d(x)
we can get:
eax sin(bx) Z aeax sin(bx)
−
dx
b
b
eax sin(bx) a Z ax
−
e sin(bx)dx
b
b
Using integration by parts again:
Z
eax sin(bx) a ax − cos(bx)
− cos(bx)
− e (
) − aeax (
)dx
b
b
b
b
eax sin(bx)
a ax
a2 Z ax
I = e cos(bx)dx =
+ 2 e cos(bx) − 2 e cos(bx)dx
b
b
b
Rewriting this:
Z
Z
ax
a2 Z ax
eax
e cos(bx)dx + 2 e cos(bx)dx = 2 [b sin(bx) + a cos(bx)]
b
b
ax
= 1+
a2 Z ax
eax
e
cos(bx)dx
=
[b sin(bx) + a cos(bx)]
b2
b2
(b2 + a2 )
Z
eax cos(bx)dx = eax [b sin(bx) + a cos(bx)]
5
eax
[b sin(bx) + a cos(bx)]
a2 + b 2
We can apply this general form back to the original integral.
I=
Z
eax cos(bx)dx =
Z
eu cos(u)du
For this case:
Z
a=1 b=1
ex
u
[sin(x) + cos(x)]
e cos(u)du =
1+1
ex (sin(x) + cos(x))
=
2
Practice Problem 39
Z
θ3 cos(θ2 )dθ
First we start with subsitution:
Let : θ2 = x
dx dx
2θ =
= θdθ
dθ
2
Z
1Z
2
2
x cos(x)dx
θ cos(θ )θdθ =
2
Then we can use integration by parts:
Z
x cos(x)dx
Let : u = x dv = cos(x)dx
du = dx v = sin(x)
x cos(x)dx = x sin(x) −
Z
sin(x)dx
x sin(x) − cos(x) + C
Now we can substitute this back:
1Z
1
x cos(x)dx = x sin(x) − cos(x) + C
2
2
1
= θ2 sin(θ2 ) − cos(θ2 ) + C
2
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Practice Problem 44
Z
3
x 2 ln(x)dx
3
Let : u = ln(x) dv = x 2 dx
5
x2
1
du = dx v = 5
x
2
5
5
Z
3
2
x ln(x)dx =
x 2 ln(x)
5
2
−
Z
1 x2
dx
x 52
5
2
2Z 3
ln(x)x 2 −
x 2 dx
5
5
5
5
2 x2
2
ln(x)x 2 − 5 dx
5
5 2
If any errors are found, please contact me at [email protected]
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