8.2 Integration by Parts
Derivative of a Product
What is the formula for the derivative of a product?
Derivative of a Product
What is the formula for the derivative of a product?
d
[f (x)g(x)] = f 0 (x)g(x) + g0 (x)f (x)
dx
Derivative of a Product
What is the formula for the derivative of a product?
d
[f (x)g(x)] = f 0 (x)g(x) + g0 (x)f (x)
dx
If we integrate both sides and rearrange, we get
Z
0
g (x)f (x) dx =
Z
d
[f (x)g(x)] dx −
dx
Z
f 0 (x)g(x) dx
Integration by Parts
So ...
R
g0 (x)f (x)dx = f (x)g(x) −
R
f 0 (x)g(x)dx
Integration by Parts
So ...
R
g0 (x)f (x)dx = f (x)g(x) −
R
uv0 = uv −
R
R
f 0 (x)g(x)dx
vu0
where u = f (x) and v = g(x).
Integration by Parts
So ...
R
g0 (x)f (x)dx = f (x)g(x) −
R
uv0 = uv −
R
R
f 0 (x)g(x)dx
vu0
where u = f (x) and v = g(x).
Integration by parts is the product rule “in reverse” in the same
way substitution is related to the chain rule.
Choosing u and v0
1
We want to get simpler with v and u0 , or at least less
complicated.
Choosing u and v0
1
2
We want to get simpler with v and u0 , or at least less
complicated.
We need to be able to find v.
First Example
Example
Integrate
Z
xe3x dx
First Example
Example
Integrate
Z
xe3x dx
This cannot be solved by any method we have seen before.
First Example
Example
Integrate
Z
xe3x dx
This cannot be solved by any method we have seen before.
The setup:
u=
dv =
du =
v=
First Example
Example
Integrate
Z
xe3x dx
This cannot be solved by any method we have seen before.
The setup:
u=
dv =
What is a good choice for u?
du =
v=
First Example
If we choose u = e3x , we do not simplify our situation at all and
2
actually make it more complicated with dv = x ⇒ v = x2 .
First Example
If we choose u = e3x , we do not simplify our situation at all and
2
actually make it more complicated with dv = x ⇒ v = x2 .
u=x
dv = e3x dx
we do make things simpler.
du = dx
v = 13 e3x
First Example
If we choose u = e3x , we do not simplify our situation at all and
2
actually make it more complicated with dv = x ⇒ v = x2 .
u=x
dv = e3x dx
du = dx
v = 13 e3x
we do make things simpler.
We can now rewrite
Z
xe3x dx =
x 3x 1
e −
3
3
Z
e3x dx
First Example
If we choose u = e3x , we do not simplify our situation at all and
2
actually make it more complicated with dv = x ⇒ v = x2 .
u=x
dv = e3x dx
du = dx
v = 13 e3x
we do make things simpler.
We can now rewrite
Z
x 3x 1
e −
e3x dx
3
3
1
x
= e3x − e3x + C
3
9
xe3x dx =
Z
Hint
So, using a power function is a good choice for u since we can
take multiple derivatives and that term will “go away” at some
point.
Hint
So, using a power function is a good choice for u since we can
take multiple derivatives and that term will “go away” at some
point.
This is usually the case ... usually ...
Second Example
Example
Z
z lnz dz
Second Example
Example
Z
z lnz dz
What happens if we choose u = z?
Second Example
Example
Z
z lnz dz
What happens if we choose u = z?
dv = ln z.
Second Example
Example
Z
z lnz dz
What happens if we choose u = z?
dv = ln z.
We don’t know how to to find the antiderivative of ln z yet, so
this is not a good choice here.
Second Example
u=
Second Example
u = ln z
Second Example
u = ln z
du =
Second Example
u = ln z
1
du = dz
z
Second Example
u = ln z
1
du = dz
z
dv =
Second Example
u = ln z
1
du = dz
z
dv = z dz
Second Example
u = ln z
1
du = dz
z
dv = z dz
v=
Second Example
u = ln z
1
du = dz
z
dv = z dz
z2
v=
2
Second Example
dv = z dz
z2
v=
2
u = ln z
1
du = dz
z
Z
z lnz dz =
z2 lnz
−
2
Z
z2 1
dz
2 z
Second Example
dv = z dz
z2
v=
2
u = ln z
1
du = dz
z
Z
z lnz dz =
z2 lnz
z2 1
−
dz
2
2 z
Z
z2 lnz
z
=
−
dz
2
2
Z
Second Example
dv = z dz
z2
v=
2
u = ln z
1
du = dz
z
Z
z lnz dz =
z2 lnz
z2 1
−
dz
2
2 z
Z
z2 lnz
z
=
−
dz
2
2
z2 lnz z2
=
− +C
2
4
Z
ln x
Example
Z
ln x dx
ln x
Example
Z
Our first IBP trick ...
u = ln x
ln x dx
ln x
Example
Z
Our first IBP trick ...
u = ln x
1
du = dx
x
ln x dx
ln x
Example
Z
ln x dx
Our first IBP trick ...
u = ln x
1
du = dx
x
dv = dx
ln x
Example
Z
ln x dx
Our first IBP trick ...
u = ln x
1
du = dx
x
dv = dx
v=x
ln x
Example
Z
ln x dx
Our first IBP trick ...
u = ln x
1
du = dx
x
Z
ln x dx = x ln x −
dv = dx
v=x
Z
dx
ln x
Example
Z
ln x dx
Our first IBP trick ...
u = ln x
1
du = dx
x
Z
dv = dx
v=x
ln x dx = x ln x −
Z
dx
= xln x − x + C
Third Example
Example
Z
x cos x dx
Third Example
Example
Z
u=
x cos x dx
Third Example
Example
Z
u=x
x cos x dx
Third Example
Example
Z
u=x
du =
x cos x dx
Third Example
Example
Z
u=x
du = dx
x cos x dx
Third Example
Example
Z
u=x
du = dx
x cos x dx
dv =
Third Example
Example
Z
u=x
du = dx
x cos x dx
dv = cos x dx
Third Example
Example
Z
u=x
du = dx
x cos x dx
dv = cos x dx
v=
Third Example
Example
Z
x cos x dx
u=x
dv = cos x dx
du = dx
v = sin x
Third Example
Example
Z
x cos x dx
u=x
dv = cos x dx
du = dx
v = sin x
Z
x cos x dx =
Third Example
Example
Z
x cos x dx
u=x
dv = cos x dx
du = dx
v = sin x
Z
x cos x dx = x sin x −
Z
sin x dx
Third Example
Example
Z
x cos x dx
u=x
dv = cos x dx
du = dx
v = sin x
Z
x cos x dx = x sin x −
Z
sin x dx
= x sin x + cos x + C
Example 4
Example
Z
x2 cos x dx
Example 4
Example
Z
u=
x2 cos x dx
Example 4
Example
Z
u = x2
x2 cos x dx
Example 4
Example
Z
u = x2
du =
x2 cos x dx
Example 4
Example
Z
u = x2
du = 2xdx
x2 cos x dx
Example 4
Example
Z
u = x2
du = 2xdx
x2 cos x dx
dv =
Example 4
Example
Z
u = x2
du = 2xdx
x2 cos x dx
dv = cos x dx
Example 4
Example
Z
u = x2
du = 2xdx
x2 cos x dx
dv = cos x dx
v=
Example 4
Example
Z
x2 cos x dx
u = x2
dv = cos x dx
du = 2xdx
v = sin x
Example 4
Example
Z
x2 cos x dx
u = x2
dv = cos x dx
du = 2xdx
v = sin x
Z
x2 cos x dx =
Example 4
Example
Z
x2 cos x dx
u = x2
dv = cos x dx
du = 2xdx
v = sin x
Z
2
2
x cos x dx = x sin x − 2
Z
x sin x dx
Example 4
2
x sin x − 2
Z
x sin x dx
Example 4
2
x sin x − 2
u=
Z
x sin x dx
Example 4
2
x sin x − 2
u=x
Z
x sin x dx
Example 4
2
x sin x − 2
u=x
du =
Z
x sin x dx
Example 4
2
x sin x − 2
u=x
du = dx
Z
x sin x dx
Example 4
2
x sin x − 2
u=x
du = dx
Z
x sin x dx
dv =
Example 4
2
x sin x − 2
u=x
du = dx
Z
x sin x dx
dv = sin x dx
Example 4
2
x sin x − 2
u=x
du = dx
Z
x sin x dx
dv = sin x dx
v=
Example 4
2
x sin x − 2
Z
x sin x dx
u=x
dv = sin x dx
du = dx
v = − cos x
Example 4
2
x sin x − 2
2
Z
x sin x dx
u=x
dv = sin x dx
du = dx
v = − cos x
x sin x − 2
Z
x sin x dx =
Example 4
2
x sin x − 2
2
Z
x sin x dx
u=x
dv = sin x dx
du = dx
v = − cos x
x sin x − 2
Z
2
x sin x dx = x sin x − 2 −x cos x −
Z
−cos x dx
Example 4
2
x sin x − 2
2
Z
x sin x dx
u=x
dv = sin x dx
du = dx
v = − cos x
x sin x − 2
Z
2
x sin x dx = x sin x − 2 −x cos x −
Z
−cos x dx
= x2 sin x + 2x cos x − 2sin x + C
Example 5
Example
Z
t2 e5t dt
Example 5
Example
Z
u=
t2 e5t dt
Example 5
Example
Z
u = t2
t2 e5t dt
Example 5
Example
Z
u = t2
du =
t2 e5t dt
Example 5
Example
Z
u = t2
du = 2t dt
t2 e5t dt
Example 5
Example
Z
u = t2
du = 2t dt
t2 e5t dt
dv =
Example 5
Example
Z
u = t2
du = 2t dt
t2 e5t dt
dv = e5t dt
Example 5
Example
Z
u = t2
du = 2t dt
t2 e5t dt
dv = e5t dt
v=
Example 5
Example
Z
u = t2
du = 2t dt
t2 e5t dt
dv = e5t dt
1
v = e5t
5
Example 5
Example
Z
u = t2
du = 2t dt
Z
t2 e5t dt =
t2 e5t dt
dv = e5t dt
1
v = e5t
5
Example 5
Example
Z
t2 e5t dt
dv = e5t dt
1
v = e5t
5
u = t2
du = 2t dt
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
Z
t e5t dt
Example 5
Z
t2 e5t dt =
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
Z
t e5t dt
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
u=
2 5t
Z
t e5t dt
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
Z
t e5t dt
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
du =
Z
t e5t dt
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
du = dt
Z
t e5t dt
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
du = dt
Z
t e5t dt
dv =
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
du = dt
Z
t e5t dt
dv = e5t dt
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
du = dt
Z
t e5t dt
dv = e5t dt
v=
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
du = dt
Z
t e5t dt
dv = e5t dt
1
v = e5t
5
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
u=t
du = dt
Z
t2 e5t dt =
Z
t e5t dt
dv = e5t dt
1
v = e5t
5
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
Z
dv = e5t dt
1
v = e5t
5
u=t
du = dt
Z
t2 e5t dt =
t e5t dt
t2 5t 2
e −
5
5
Z
t e5t dt
Example 5
Z
t2 5t 2
t e dt =
e −
5
5
2 5t
Z
dv = e5t dt
1
v = e5t
5
u=t
du = dt
Z
t e5t dt
t2 5t 2
e −
t e5t dt
5
5
t2
2t
2 5t
= e5t − e5t +
e +C
5
25
125
t2 e5t dt =
Z
Example 6
Example
Z
x2 ln x dx =
Example 6
Example
Z
u=
x2 ln x dx =
Example 6
Example
Z
u = ln x
du =
x2 ln x dx =
Example 6
Example
Z
u = ln x
1
du = dx
x
x2 ln x dx =
Example 6
Example
Z
u = ln x
1
du = dx
x
x2 ln x dx =
dv =
Example 6
Example
Z
u = ln x
1
du = dx
x
x2 ln x dx =
dv = x2 dx
Example 6
Example
Z
u = ln x
1
du = dx
x
x2 ln x dx =
dv = x2 dx
v=
Example 6
Example
Z
u = ln x
1
du = dx
x
x2 ln x dx =
dv = x2 dx
x3
v=
3
Example 6
Example
Z
u = ln x
1
du = dx
x
Z
x2 ln x dx =
x2 ln x dx =
dv = x2 dx
x3
v=
3
Example 6
Example
Z
x2 ln x dx =
dv = x2 dx
x3
v=
3
u = ln x
1
du = dx
x
Z
x2 ln x dx =
x3
1
ln x −
3
3
Z
x2 dx
Example 6
Example
Z
u = ln x
1
du = dx
x
Z
x2 ln x dx =
x2 ln x dx =
dv = x2 dx
x3
v=
3
x3
1
ln x −
x2 dx
3
3
x3
x3
=
ln x −
+C
3
9
Z
Example 7
Example
Z
y
p
y + 3 dy
Example 7
Example
Z
u=
y
p
y + 3 dy
Example 7
Example
Z
u=y
du =
y
p
y + 3 dy
Example 7
Example
Z
u=y
du = dy
y
p
y + 3 dy
Example 7
Example
Z
u=y
du = dy
y
p
y + 3 dy
dv =
Example 7
Example
Z
u=y
du = dy
y
p
y + 3 dy
dv =
p
y + 3 dy
Example 7
Example
Z
y
p
y + 3 dy
u=y
dv =
du = dy
v=
p
y + 3 dy
Example 7
Example
Z
y
p
y + 3 dy
u=y
dv =
du = dy
v=
p
y + 3 dy
3
2
(y + 3) 2
3
Example 7
Example
Z
Z
y
p
y + 3 dy
u=y
dv =
du = dy
v=
y
p
y + 3 dy =
p
y + 3 dy
3
2
(y + 3) 2
3
Example 7
Example
Z
Z
y
p
y + 3 dy
u=y
dv =
du = dy
v=
y
p
y + 3 dy =
3
2
2
y(y + 3) 2 −
3
3
Z
p
y + 3 dy
3
2
(y + 3) 2
3
3
(y + 3) 2 dy
Example 7
Example
Z
Z
y
p
y + 3 dy
u=y
dv =
du = dy
v=
y
p
p
y + 3 dy
3
2
(y + 3) 2
3
3
3
2
2
y(y + 3) 2 −
(y + 3) 2 dy
3
3
3
5
2
4
= y(y + 3) 2 − (y + 3) 2 + C
3
15
y + 3 dy =
Z
Example 8
Example
Z
√
(x + 2) 2 + 4x dx
Example 8
Example
Z
u=
√
(x + 2) 2 + 4x dx
Example 8
Example
Z
u = x+2
du =
√
(x + 2) 2 + 4x dx
Example 8
Example
Z
u = x+2
du = dx
√
(x + 2) 2 + 4x dx
Example 8
Example
Z
u = x+2
du = dx
√
(x + 2) 2 + 4x dx
dv =
Example 8
Example
Z
u = x+2
du = dx
√
(x + 2) 2 + 4x dx
dv =
√
2 + 4x dx
Example 8
Example
Z
u = x+2
du = dx
√
(x + 2) 2 + 4x dx
dv =
v=
√
2 + 4x dx
Example 8
Example
Z
u = x+2
du = dx
√
(x + 2) 2 + 4x dx
dv =
√
2 + 4x dx
3
1
v = (2 + 4x) 2
6
Example 8
Example
Z
u = x+2
du = dx
Z
√
(x + 2) 2 + 4x dx =
√
(x + 2) 2 + 4x dx
dv =
√
2 + 4x dx
3
1
v = (2 + 4x) 2
6
Example 8
Example
Z
u = x+2
du = dx
Z
√
(x + 2) 2 + 4x dx
dv =
√
2 + 4x dx
3
1
v = (2 + 4x) 2
6
Z
√
3
3
1
1
(2 + 4x) 2 dx
(x + 2) 2 + 4x dx = (x + 2)(2 + 4x) 2 −
6
6
Example 8
Example
Z
u = x+2
du = dx
Z
√
(x + 2) 2 + 4x dx
dv =
√
2 + 4x dx
3
1
v = (2 + 4x) 2
6
Z
√
3
3
1
1
(2 + 4x) 2 dx
(x + 2) 2 + 4x dx = (x + 2)(2 + 4x) 2 −
6
6
3
5
1
1
= (x + 2)(2 + 4x) 2 − (2 + 4x) 2 + C
6
60
Example 9
Example
Z
x
dx
ex
Example 9
Example
Z
u=
x
dx
ex
Example 9
Example
Z
u=x
du =
x
dx
ex
Example 9
Example
Z
u=x
du = dx
x
dx
ex
Example 9
Example
Z
u=x
du = dx
x
dx
ex
dv =
Example 9
Example
Z
u=x
du = dx
x
dx
ex
dv = e−x dx
Example 9
Example
Z
u=x
du = dx
x
dx
ex
dv = e−x dx
v=
Example 9
Example
Z
x
dx
ex
u=x
dv = e−x dx
du = dx
v = − e−x
Example 9
Example
Z
x
dx
ex
u=x
dv = e−x dx
du = dx
v = − e−x
Z
x
dx =
ex
Example 9
Example
Z
x
dx
ex
u=x
dv = e−x dx
du = dx
v = − e−x
Z
x
dx = − x e−x −
ex
Z
−e−x dx
Example 9
Example
Z
x
dx
ex
u=x
dv = e−x dx
du = dx
v = − e−x
Z
x
dx = − x e−x − −e−x dx
ex
= − x e−x − e−x + C
Z
Example 10
Example
Z
tan−1 7x dx
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
u = tan−1 7x
du =
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
u = tan−1 7x
7
du =
dx
1 + 49x2
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
u = tan−1 7x
7
du =
dx
1 + 49x2
dv = dx
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
u = tan−1 7x
7
du =
dx
1 + 49x2
dv = dx
v=
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
u = tan−1 7x
7
du =
dx
1 + 49x2
dv = dx
v=x
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
u = tan−1 7x
7
du =
dx
1 + 49x2
Z
tan−1 7x dx =
dv = dx
v=x
Example 10
Example
Z
tan−1 7x dx
Anyone remember the formula for the derivative of the
arctangent function?
u = tan−1 7x
7
du =
dx
1 + 49x2
Z
tan
−1
7x dx = xtan
dv = dx
v=x
−1
7x − 7
Z
x dx
1 + 49x2
Example 10
Z
tan−1 7 x dx =
Example 10
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
Example 10
Z
tan
−1
7 x dx = x tan
−1
We can now apply substitution.
7x−7
Z
x dx
1 + 49x2
Example 10
Z
tan
−1
7 x dx = x tan
−1
We can now apply substitution.
w = 1 + 49x2
7x−7
Z
x dx
1 + 49x2
Example 10
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
We can now apply substitution.
w = 1 + 49x2
dw = 98x dx
Example 10
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
We can now apply substitution.
w = 1 + 49x2
dw = 98x dx
1
dw = x dx
98
Example 10
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
We can now apply substitution.
dw = 98x dx
1
dw = x dx
98
w = 1 + 49x2
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
Example 10
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
We can now apply substitution.
dw = 98x dx
1
dw = x dx
98
w = 1 + 49x2
Z
tan
−1
x dx
1 + 49x2
Z
dw
1
−1
= x tan 7 x −
14
w
7 x dx = x tan
−1
7x−7
Z
Example 10
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
We can now apply substitution.
dw = 98x dx
1
dw = x dx
98
w = 1 + 49x2
Z
tan
−1
x dx
1 + 49x2
Z
dw
1
−1
= x tan 7 x −
14
w
1
= x tan−1 7 x −
ln |w| + C
14
7 x dx = x tan
−1
7x−7
Z
Example 10
Z
tan
−1
7 x dx = x tan
−1
7x−7
Z
x dx
1 + 49x2
We can now apply substitution.
dw = 98x dx
1
dw = x dx
98
w = 1 + 49x2
Z
tan
x dx
1 + 49x2
Z
dw
1
−1
= x tan 7 x −
14
w
1
= x tan−1 7 x −
ln |w| + C
14
1
= x tan−1 7 x −
ln |1 + 49x2 | + C
14
−1
7 x dx = x tan
−1
7x−7
Z
Example 11
Example
Z
sin2 x dx
Example 11
Example
Z
sin2 x dx
Our first instinct would probably be to try using
Example 11
Example
Z
sin2 x dx
Our first instinct would probably be to try using u = sin x and
dv = cos x dx.
Example 11
Example
Z
sin2 x dx
Our first instinct would probably be to try using u = sin x and
dv = cos x dx.
When we do we end up with −sin x cos x +
R
cos2 x dx.
Example 11
Example
Z
sin2 x dx
Our first instinct would probably be to try using u = sin x and
dv = cos x dx.
When we do we end up with −sin x cos x +
R
cos2 x dx.
Since this isn’t as far as we need to be, we may try IBP again
with u = cos x and dv = cos x dx this time.
Example 11
Example
Z
sin2 x dx
Our first instinct would probably be to try using u = sin x and
dv = cos x dx.
When we do we end up with −sin x cos x +
R
cos2 x dx.
Since this isn’t as far as we need to be, we may try IBP again
with u = cos x and dv = cos x dx this time.
Then, we’d have = −sin x cos x + sin x cos x +
which gets us exactly nowhere.
R
sin2 x dx,
Example 11
But what if after the first application of IBP we tried this
instead:
Z
sin2 x dx = −sin xcos x +
Z
cos2 x dx
Example 11
But what if after the first application of IBP we tried this
instead:
Z
Z
2
sin2 x dx = −sin xcos x +
sin x dx = −sin xcos x +
Z
Z
cos2 x dx
(1 − sin2 x) dx
Example 11
But what if after the first application of IBP we tried this
instead:
Z
Z
Z
sin2 x dx = −sin xcos x +
2
sin x dx = −sin xcos x +
sin2 x dx = −sin xcos x +
Z
Z
Z
cos2 x dx
(1 − sin2 x) dx
dx −
Z
sin2 x dx
Example 11
But what if after the first application of IBP we tried this
instead:
Z
Z
Z
sin2 x dx = −sin xcos x +
2
sin x dx = −sin xcos x +
sin2 x dx = −sin xcos x +
⇒2
Z
Z
Z
Z
cos2 x dx
(1 − sin2 x) dx
dx −
Z
sin2 x dx
sin2 x dx = −sin x cos x + x + C
Example 11
But what if after the first application of IBP we tried this
instead:
Z
Z
Z
sin2 x dx = −sin xcos x +
2
sin x dx = −sin xcos x +
sin2 x dx = −sin xcos x +
⇒2
⇒
Z
Z
Z
Z
Z
cos2 x dx
(1 − sin2 x) dx
dx −
Z
sin2 x dx
sin2 x dx = −sin x cos x + x + C
1
x
sin2 x dx = − sin x cos x + + C
2
2
Example 12
Example
Z 1
0
sin−1 x dx
Example 12
Example
Z 1
0
sin−1 x dx
Does anyone remember the derivative of sin−1 x?
Example 12
Example
Z 1
0
sin−1 x dx
Does anyone remember the derivative of sin−1 x?
u = sin−1 x
du =
Example 12
Example
Z 1
0
sin−1 x dx
Does anyone remember the derivative of sin−1 x?
u = sin−1 x
dx
du = √
1 − x2
Example 12
Example
Z 1
0
sin−1 x dx
Does anyone remember the derivative of sin−1 x?
u = sin−1 x
dx
du = √
1 − x2
dv = dx
Example 12
Example
Z 1
0
sin−1 x dx
Does anyone remember the derivative of sin−1 x?
u = sin−1 x
dx
du = √
1 − x2
dv = dx
v=
Example 12
Example
Z 1
0
sin−1 x dx
Does anyone remember the derivative of sin−1 x?
u = sin−1 x
dx
du = √
1 − x2
dv = dx
v=x
Example 12
Example
Z 1
0
sin−1 x dx
Does anyone remember the derivative of sin−1 x?
u = sin−1 x
dx
du = √
1 − x2
dv = dx
v=x
This gives
Z 1
0
1 Z
sin−1 x dx = x sin−1 x −
0
1
0
x dx
√
1 − x2
Example 12
Z 1
0
1 Z
sin−1 x dx = x sin−1 x −
0
1
0
x dx
√
1 − x2
Example 12
Z 1
0
1 Z
sin−1 x dx = x sin−1 x −
0
=
π
−
2
1
0
Z 1
0
x dx
√
1 − x2
x dx
√
1 − x2
Example 12
Z 1
0
1 Z
sin−1 x dx = x sin−1 x −
0
=
Now we can use substitution.
w = 1 − x2
π
−
2
1
0
Z 1
0
x dx
√
1 − x2
x dx
√
1 − x2
Example 12
Z 1
0
1 Z
sin−1 x dx = x sin−1 x −
0
=
π
−
2
Now we can use substitution.
w = 1 − x2 ⇒ dw = −2x dx
1
0
Z 1
0
x dx
√
1 − x2
x dx
√
1 − x2
Example 12
Z 1
0
1 Z
sin−1 x dx = x sin−1 x −
0
=
π
−
2
1
0
Z 1
0
x dx
√
1 − x2
x dx
√
1 − x2
Now we can use substitution.
1
w = 1 − x2 ⇒ dw = −2x dx ⇒ − dw = x dx
2
Example 12
Z 1
0
sin
−1
π 1
x dx = +
2
2
Z x=1
dw
√
x=0
w
Example 12
Z 1
0
sin
−1
π 1 x=1 dw
√
x dx = +
2
2 x=0
w
x
=
1
π 1
1
= + (2)w 2 2
2
x=0
Z
Example 12
Z 1
0
sin
−1
π 1 x=1 dw
√
x dx = +
2
2 x=0
w
x
=
1
π 1
1
= + (2)w 2 2
2
x=0
1 1
π
= + ( 1 − x2 ) 2 2
0
Z
Example 12
Z 1
0
sin
−1
π 1 x=1 dw
√
x dx = +
2
2 x=0
w
x
=
1
π 1
1
= + (2)w 2 2
2
x=0
1 1
π
= + ( 1 − x2 ) 2 2
0
π
= −1
2
Z
Example 13
Example
Z
x5 cos(x3 ) dx
Example 13
Example
Z
x5 cos(x3 ) dx
We can begin with an old substitution trick.
Example 13
Example
Z
x5 cos(x3 ) dx
We can begin with an old substitution trick.
w = x3
Example 13
Example
Z
x5 cos(x3 ) dx
We can begin with an old substitution trick.
w = x3
dw = 3x2 dx
Example 13
Example
Z
x5 cos(x3 ) dx
We can begin with an old substitution trick.
w = x3
dw = 3x2 dx
1
dw = x2 dx
3
Example 13
Example
Z
x5 cos(x3 ) dx
We can begin with an old substitution trick.
w = x3
dw = 3x2 dx
1
dw = x2 dx
3
This gives
1
3
Now we can apply IBP.
Z
w cos w dw
Example 14
1
3
Z
w cos w dw
Example 14
1
3
u=
Z
w cos w dw
Example 14
1
3
u=w
du =
Z
w cos w dw
Example 14
1
3
u=w
du = dw
Z
w cos w dw
Example 14
1
3
u=w
du = dw
Z
w cos w dw
dv = cos w dw
Example 14
1
3
u=w
du = dw
Z
w cos w dw
dv = cos w dw
v=
Example 14
1
3
Z
w cos w dw
u=w
dv = cos w dw
du = dw
v = sin w
Example 14
1
3
1
3
Z
w cos w dw
u=w
dv = cos w dw
du = dw
v = sin w
Z
w cos w dw =
Example 14
1
3
1
3
Z
w cos w dw
u=w
dv = cos w dw
du = dw
v = sin w
Z
w cos w dw =
1
1
w sin w −
3
3
Z
sin w dw
Example 14
1
3
1
3
Z
w cos w dw
u=w
dv = cos w dw
du = dw
v = sin w
Z
1
1
w sin w −
sin w dw
3
3
1
1
= w sin w + cos w + C
3
3
w cos w dw =
Z
Example 14
1
3
1
3
Z
w cos w dw
u=w
dv = cos w dw
du = dw
v = sin w
Z
1
1
w sin w −
sin w dw
3
3
1
1
= w sin w + cos w + C
3
3
1 3
=
x sin x3 + cos x3 + C
3
w cos w dw =
Z
Example 14
Example
Z 5
0
ln(1 + t) dt
Example 14
Example
Z 5
0
ln(1 + t) dt
We first could use substitution with w = t + 1. Then, we could
use IBP, but since we already found a formula for lnx, we could
apply that and just evaluate.
Example 14
Example
Z 5
0
ln(1 + t) dt
We first could use substitution with w = t + 1. Then, we could
use IBP, but since we already found a formula for lnx, we could
apply that and just evaluate.
Z 5
0
5
ln(1 + t) dt = (1 + t)ln(1 + t) − (t + 1) = 6 ln 6 − 5
0
Example 15
Example
Z √
x ln x dx
Example 15
Example
Z √
u=
x ln x dx
Example 15
Example
Z √
u = ln x
x ln x dx
Example 15
Example
Z √
u = ln x
dx
du =
x
x ln x dx
Example 15
Example
Z √
u = ln x
dx
du =
x
x ln x dx
dv =
√
x dx
Example 15
Example
Z √
u = ln x
dx
du =
x
x ln x dx
dv =
√
x dx
2 3
v = x2
3
Example 15
Example
Z √
u = ln x
dx
du =
x
Z √
x ln x dx =
x ln x dx
dv =
√
x dx
2 3
v = x2
3
Example 15
Example
Z √
u = ln x
dx
du =
x
Z √
x ln x dx =
x ln x dx
dv =
√
x dx
2 3
v = x2
3
3
2
2
(ln x) x 2 −
3
3
Z
1
x 2 dx
Example 15
Example
Z √
dv =
√
x dx
2 3
v = x2
3
u = ln x
dx
du =
x
Z √
x ln x dx
1
3
2
2
(ln x) x 2 −
x 2 dx
3
3
3
2
4 3
= (ln x) x 2 − x 2 + C
3
9
x ln x dx =
Z
Example 16
Example
Z
ex cosx dx
Example 16
Example
Z
ex cosx dx
Here we apply IBP twice, both times with u = ex . Then we do
some algebra like we did before with the integral of sin2 x.
Example 16
Example
Z
ex cosx dx
Here we apply IBP twice, both times with u = ex . Then we do
some algebra like we did before with the integral of sin2 x.
Z
ex cos x dx =
1
1 x
e sin x + ex cos x + C
2
2
Example 17
Example
Z
cos2 (3α + 1) dα
Example 17
Example
Z
cos2 (3α + 1) dα
Similar to the sin2 x with u = dv = cos(3α + 1).
Example 17
Example
Z
cos2 (3α + 1) dα
Similar to the sin2 x with u = dv = cos(3α + 1).
We end up with
Z
cos2 (3α + 1) dα =
1
α
sin(3α + 1)cos(3α + 1) + + c
6
2
Where We Are At
So, at this point, we have the following techniques that we
could apply:
substitution
Z
√
4x
2x2 + 1
dx
Where We Are At
So, at this point, we have the following techniques that we
could apply:
substitution
Z
√
substitution
Z
4x
2x2 + 1
dx
etanx
dx
cos2 x
Where We Are At
So, at this point, we have the following techniques that we
could apply:
substitution
Z
√
substitution
4x
2x2 + 1
dx
Z
etanx
dx
cos2 x
Z
x2
dx
4 + x2
long division
What We Could Use
completing the square
Z
x2
dx
+ 6x + 14
dx
(x + 3)2 + 5
1
−1 x + 3
√
= √ tan
+c
5
5
=
Z
What We Could Use
completing the square
Z
x2
dx
+ 6x + 14
dx
(x + 3)2 + 5
1
−1 x + 3
√
= √ tan
+c
5
5
=
Z
partial fraction decomposition
Z
dx
x(x + 1)
What We Could Use
completing the square
Z
x2
dx
+ 6x + 14
dx
(x + 3)2 + 5
1
−1 x + 3
√
= √ tan
+c
5
5
=
Z
partial fraction decomposition
Z
dx
x(x + 1)
What if none of these techniques work? We then will turn our
attention integration using tables.