Strategy for using integration by parts
Recall the integration by parts formula:
Z
Z
u dv = uv − v du.
Strategy for using integration by parts
Recall the integration by parts formula:
Z
Z
u dv = uv − v du.
To apply this formula we must choose dv so that we
can integrate it!
Strategy for using integration by parts
Recall the integration by parts formula:
Z
Z
u dv = uv − v du.
To apply this formula we must choose dv so that we
can integrate it! Frequently, we choose u so that the
derivative of u is simpler than u.
Strategy for using integration by parts
Recall the integration by parts formula:
Z
Z
u dv = uv − v du.
To apply this formula we must choose dv so that we
can integrate it! Frequently, we choose u so that the
derivative of u is simpler than u. If both properties
hold, then you have made the correct choice.
Examples using strategy:
Z
1
xe x dx :
R
u dv = uv −
R
v du
Examples using strategy:
Z
1
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
Examples using strategy:
Z
1
Z
2
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt :
Examples using strategy:
Z
1
Z
2
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt : Choose u = t2 and dv = et dt
Examples using strategy:
Z
1
Z
2
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt : Choose u = t2 and dv = et dt
Z
3
R
ln x dx :
Examples using strategy:
Z
1
Z
2
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt : Choose u = t2 and dv = et dt
Z
3
ln x dx : Choose u = ln x and dv = dx
Examples using strategy:
Z
1
Z
2
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt : Choose u = t2 and dv = et dt
Z
ln x dx : Choose u = ln x and dv = dx
3
Z
4
x sin x dx :
Examples using strategy:
Z
1
Z
2
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt : Choose u = t2 and dv = et dt
Z
ln x dx : Choose u = ln x and dv = dx
3
Z
4
x sin x dx : u = x and dv = sin x dx
Examples using strategy:
Z
1
Z
2
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt : Choose u = t2 and dv = et dt
Z
ln x dx : Choose u = ln x and dv = dx
3
Z
x sin x dx : u = x and dv = sin x dx
4
Z
5
x 2 sin 2x dx :
Examples using strategy:
Z
1
Z
2
R
u dv = uv −
R
v du
xe x dx : Choose u = x and dv = ex dx
t 2 e t dt : Choose u = t2 and dv = et dt
Z
ln x dx : Choose u = ln x and dv = dx
3
Z
x sin x dx : u = x and dv = sin x dx
4
Z
5
x 2 sin 2x dx : u = x2 and dv = sin 2x dx
R
u dv = uv −
R
v du
R
Example Find xe x dx.
R
u dv = uv −
R
v du
R
Example Find xe x dx.
Solution Let
u=x
dv = ex dx.
R
u dv = uv −
R
v du
R
Example Find xe x dx.
Solution Let
u=x
dv = ex dx.
Then
du = dx
v = ex .
R
u dv = uv −
R
v du
R
Example Find xe x dx.
Solution Let
u=x
dv = ex dx.
Then
du = dx
v = ex .
Integrating by parts gives
Z
Z
x
x
xe dx = xe − e x dx
R
u dv = uv −
R
v du
R
Example Find xe x dx.
Solution Let
u=x
dv = ex dx.
Then
du = dx
v = ex .
Integrating by parts gives
Z
Z
x
x
xe dx = xe − e x dx = xe x − e x + C .
R
u dv = uv −
R
v du
R
Example Evaluate ln x dx.
R
u dv = uv −
R
v du
R
Example Evaluate ln x dx.
Solution Let
u = ln x
dv = dx.
R
u dv = uv −
R
v du
R
Example Evaluate ln x dx.
Solution Let
u = ln x
Then
1
du = dx
x
dv = dx.
v = x.
R
u dv = uv −
R
v du
R
Example Evaluate ln x dx.
Solution Let
u = ln x
Then
dv = dx.
1
du = dx v = x.
x
Integrating by parts, we get
Z
Z
dx
ln x dx = x ln x − x
x
R
u dv = uv −
R
v du
R
Example Evaluate ln x dx.
Solution Let
u = ln x
Then
dv = dx.
1
du = dx v = x.
x
Integrating by parts, we get
Z
Z
dx
ln x dx = x ln x − x
x
Z
= x ln x −
dx = x ln x − x + C .
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt
v = et .
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt v = et . R
R
Integration by parts gives t 2 e t dt = t 2 e t − 2 te t dt
(1)
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt v = et . R
R
Integration by parts gives t 2 e t dt = t 2 e t − 2 te t dt
Using integration by parts a second time,
(1)
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt v = et . R
R
Integration by parts gives t 2 e t dt = t 2 e t − 2 te t dt (1)
Using integration by parts a second time, this time with
u=t
dv = et dt.
Then
du = dt,
v = et ,
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt v = et . R
R
Integration by parts gives t 2 e t dt = t 2 e t − 2 te t dt (1)
Using integration by parts a second time, this time with
u=t
dv = et dt.
Then
du = dt,
v = et ,
and
R t
R
te dt = te t − e t dt = te t − e t + C .
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt v = et . R
R
Integration by parts gives t 2 e t dt = t 2 e t − 2 te t dt (1)
Using integration by parts a second time, this time with
u=t
dv = et dt.
Then
du = dt,
v = et ,
and
R t
R
te dt = te t − e t dt = te t − e t + C . Putting this in
Equation (1), we get
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt v = et . R
R
Integration by parts gives t 2 e t dt = t 2 e t − 2 te t dt (1)
Using integration by parts a second time, this time with
u=t
dv = et dt.
Then
du = dt,
v = et ,
and
R t
R
te dt = te t − e t dt = te t − e t + C . Putting this in
Equation (1), we get
Z
Z
2 t
2 t
t e dt = t e − 2 te t dt = t 2 e t − 2(te t − e t + C )
R
u dv = uv −
R
v du
R
Example Find t 2e t dt.
Solution Let u = t2 dv = et dt
Then du = 2tdt v = et . R
R
Integration by parts gives t 2 e t dt = t 2 e t − 2 te t dt (1)
Using integration by parts a second time, this time with
u=t
dv = et dt.
Then
du = dt,
v = et ,
and
R t
R
te dt = te t − e t dt = te t − e t + C . Putting this in
Equation (1), we get
Z
Z
2 t
2 t
t e dt = t e − 2 te t dt = t 2 e t − 2(te t − e t + C )
= t 2 e t − 2te t + 2e t + C1 whereC1 = −2C .
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice.
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x,
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
v = sin x,
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
We put Equation (3) into Equation (2)
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
we get
We
into Equation (2)
R xput Equation (3)
R and
x
x
x
e sin x dx = −e cos x + e sin x − e sin x dx.
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
we get
We
into Equation (2)
R xput Equation (3)
R and
x
x
x
e sin x dx = −e cos x + e sin x − e sin x dx. This can
be regarded as an equation to be solved for the unknown
integral.
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
we get
We
into Equation (2)
R xput Equation (3)
R and
x
x
x
e sin x dx = −e cos x + e sin x − e sin x dx. This can
be regarded as anR equation to be solved for the unknown
integral. Adding e x sin x dx to both sides, we obtain
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
we get
We
into Equation (2)
R xput Equation (3)
R and
x
x
x
e sin x dx = −e cos x + e sin x − e sin x dx. This can
be regarded as anR equation to be solved for the unknown
x
integral.
R x Adding e sin x dx xto both sides, we obtain
2 e sin x dx = −e cos x + e sin x.
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
we get
We
into Equation (2)
R xput Equation (3)
R and
x
x
x
e sin x dx = −e cos x + e sin x − e sin x dx. This can
be regarded as anR equation to be solved for the unknown
x
integral.
R x Adding e sin x dx xto both sides, we obtain
2 e sin x dx = −e cos x + e sin x.
Dividing by 2 and adding the constant of integration, we get
R
u dv = uv −
R
v du
R
Example Evaluate e x sin x dx.
Solution Solving this integral involves integrating by parts
twice. We try choosing u = ex and dv = sin x dx. Then
du = ex dx and v = − cos x, so integration by parts gives
R x
R
e sin x dx = −e x cos x + e x cos x dx. (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,
vR = sin x, and
R
e x cos x dx = e x sin x − e x sin x dx. (3)
we get
We
into Equation (2)
R xput Equation (3)
R and
x
x
x
e sin x dx = −e cos x + e sin x − e sin x dx. This can
be regarded as anR equation to be solved for the unknown
x
integral.
R x Adding e sin x dx xto both sides, we obtain
2 e sin x dx = −e cos x + e sin x.
Dividing by 2 and adding the constant of integration, we get
R x
e sin x dx = 12 e x (sin x − cos x) + C .
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry.
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
f (x) = cos3 x
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
2
f (x) = cos3 x
f (x) = sin5 x cos2 x
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
2
3
f (x) = cos3 x
f (x) = sin5 x cos2 x
f (x) = sin2 x.
Trigonometric integrals
Trigonometric integrals are integrals of functions
f (x) that can be expressed as a product of
functions from trigonometry. For example;
1
2
3
f (x) = cos3 x
f (x) = sin5 x cos2 x
f (x) = sin2 x.
Integrating such functions involve several techniques
and strategies which we will describe today.
Aside from the most basic relations such as
sin(θ)
1
tan x =
and sec x =
, you should
cos(θ)
cos(θ)
know the following trig identities:
Aside from the most basic relations such as
sin(θ)
1
tan x =
and sec x =
, you should
cos(θ)
cos(θ)
know the following trig identities:
cos2 (θ) + sin2 (θ) = 1.
sec2 (θ) − tan2 (θ) = 1.
1 + cos(2θ)
cos2 (θ) =
2
1 − cos(2θ)
sin2 (θ) =
2
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2 (θ)−sin2 (θ) = 2 cos2 (θ)−1 = 1−2 sin2 (θ)
R
Example Evaluate cos3 x dx.
R
Example Evaluate cos3 x dx.
Solution So here we recall the formula:
sin2 x + cos2 = 1 or cos2 x = 1 − sin2 x.
R
Example Evaluate cos3 x dx.
Solution So here we recall the formula:
sin2 x + cos2 = 1 or cos2 x = 1 − sin2 x.
We then get:
Z
Z
cos3 x dx = cos2 x·cos x dx
R
Example Evaluate cos3 x dx.
Solution So here we recall the formula:
sin2 x + cos2 = 1 or cos2 x = 1 − sin2 x.
We then get:
Z
Z
Z
cos3 x dx = cos2 x·cos x dx = (1−sin2 x) cos x dx
R
Example Evaluate cos3 x dx.
Solution So here we recall the formula:
sin2 x + cos2 = 1 or cos2 x = 1 − sin2 x.
We then get:
Z
Z
Z
cos3 x dx = cos2 x·cos x dx = (1−sin2 x) cos x dx
Let u = sin x, so du = cos x dx.
R
Example Evaluate cos3 x dx.
Solution So here we recall the formula:
sin2 x + cos2 = 1 or cos2 x = 1 − sin2 x.
We then get:
Z
Z
Z
cos3 x dx = cos2 x·cos x dx = (1−sin2 x) cos x dx
Let u = sin x, so du = cos x dx. So, we get
Z
1
(1 − u 2 ) du = u − u 3 + C
3
R
Example Evaluate cos3 x dx.
Solution So here we recall the formula:
sin2 x + cos2 = 1 or cos2 x = 1 − sin2 x.
We then get:
Z
Z
Z
cos3 x dx = cos2 x·cos x dx = (1−sin2 x) cos x dx
Let u = sin x, so du = cos x dx. So, we get
Z
1
(1 − u 2 ) du = u − u 3 + C
3
= sin x −
1 3
sin x + C .
3
Rπ
Find 0 sin2 x dx.
Solution Here we use the double angle formula:
Example
sin2 x dx = 21 (1 − cos 2x).
Rπ
Find 0 sin2 x dx.
Solution Here we use the double angle formula:
Example
sin2 x dx = 21 (1 − cos 2x).
Z
0
π
sin2 x dx
Rπ
Find 0 sin2 x dx.
Solution Here we use the double angle formula:
Example
sin2 x dx = 21 (1 − cos 2x).
Z
0
π
1
sin x dx =
2
2
Z
π
(1 − cos 2x) dx
0
Rπ
Find 0 sin2 x dx.
Solution Here we use the double angle formula:
Example
sin2 x dx = 21 (1 − cos 2x).
Z
0
π
1
sin x dx =
2
2
Z
0
π
1
(1 − cos 2x) dx =
2
π
1
x − sin 2x 2
0
Rπ
Find 0 sin2 x dx.
Solution Here we use the double angle formula:
Example
sin2 x dx = 21 (1 − cos 2x).
Z
0
π
1
sin x dx =
2
2
Z
0
π
1
(1 − cos 2x) dx =
2
π
1
x − sin 2x 2
0
1
1
1
1
= (π − sin 2π) − (0 − sin 0)
2
2
2
2
Rπ
Find 0 sin2 x dx.
Solution Here we use the double angle formula:
Example
sin2 x dx = 21 (1 − cos 2x).
Z
0
π
1
sin x dx =
2
2
Z
0
π
1
(1 − cos 2x) dx =
2
π
1
x − sin 2x 2
0
1
1
1
1
1
= (π − sin 2π) − (0 − sin 0) = π.
2
2
2
2
2
Here we mentally made the substitution u = 2x when
integrating cos 2x.
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Z
Z
m
2k+1
sin x cos
x dx = sinm x(cos2 x)k cos x dx
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Z
Z
m
2k+1
sin x cos
x dx = sinm x(cos2 x)k cos x dx
Z
=
sinm x(1 − sin2 x)k cos x dx
Strategy for Evaluating
R
sinm x cosn x dx
(a) If the power of cosine is odd (n = 2k + 1), save
one cosine factor and use cos2 x = 1 − sin2 x to
express the remaining factors in terms of sine:
Z
Z
m
2k+1
sin x cos
x dx = sinm x(cos2 x)k cos x dx
Z
=
sinm x(1 − sin2 x)k cos x dx
Then substitute u = sin x.
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Z
sin2k+1 x cosn x dx =
Z
(sin2 x)k cosn x sin x dx
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Z
sin2k+1 x cosn x dx =
Z
=
Z
(sin2 x)k cosn x sin x dx
(1 − cos2 x)k cosn x sin x dx.
Strategy for Evaluating
R
sinm x cosn x dx
(b) If the power of sine is odd (m = 2k + 1), save
one sine factor and use sin2 x = 1 − cos2 x to
express the remaining factors in terms of cosine:
Z
sin2k+1 x cosn x dx =
Z
=
Z
(sin2 x)k cosn x sin x dx
(1 − cos2 x)k cosn x sin x dx.
Then substitute u = cos x.
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
sin2 x = 12 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
sin2 x = 12 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
It is sometimes helpful to use the identity
Strategy for Evaluating
R
sinm x cosn x dx
(c) If the powers of both sine and cosine are even,
use the half-angle identities
sin2 x = 12 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
It is sometimes helpful to use the identity
sin x cos x = 12 sin 2x
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Z
tanm x sec2k x dx =
Z
tanm x(sec2 x)k−1 sec2 x dx
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Z
tanm x sec2k x dx =
Z
=
Z
tanm x(sec2 x)k−1 sec2 x dx
tanm x(1 + tan2 x)k−1 sec2 x dx
Strategy for Evaluating
R
tanm x secn x dx
(a) If the power of secant is even (n = 2k, k ≥ 2),
save a factor of sec2 x and use sec2 = 1 + tan2 x
to express the remaining factors in terms of tan x:
Z
tanm x sec2k x dx =
Z
=
Z
tanm x(sec2 x)k−1 sec2 x dx
tanm x(1 + tan2 x)k−1 sec2 x dx
Then substitute u = tan x.
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Z
tan
2k+1
n
x sec x dx =
Z
(tan2 x)k secn−1 x sec x tan x dx
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Z
tan
2k+1
Z
=
n
x sec x dx =
Z
(tan2 x)k secn−1 x sec x tan x dx
(sec2 x − 1)k secn−1 x sec x tan x dx
Strategy for Evaluating
R
tanm x secn x dx
(b) If the power of tangent is odd (m = 2k + 1),
save a factor of sec x tan x and use
tan2 x = sec2 x − 1 to express the remaining
factors in terms of sec x:
Z
tan
2k+1
Z
=
n
x sec x dx =
Z
(tan2 x)k secn−1 x sec x tan x dx
(sec2 x − 1)k secn−1 x sec x tan x dx
Then substitute u = sec x.
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
The next formula can be checked by differentiating
the right hand side.
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
The next formula can be checked by differentiating
the right hand side.
Z
sec x dx = ln | sec x + tan x| + C .
Two other useful formulas
Recall that we proved the following formula is class
using integration by parts.
Z
tan x dx = ln | sec x| + C .
The next formula can be checked by differentiating
the right hand side.
Z
sec x dx = ln | sec x + tan x| + C .
Also, don’t forget that
d
dx sec x = sec x tan x.
d
dx
tan x = sec2 x and
R
Example Find tan3 x dx.
R
Example Find tan3 x dx.
Solution Here only tan x occurs, so we use
tan2 x = sec2 x − 1 to rewrite a tan2 x factor in
terms of sec2 x:
R
Example Find tan3 x dx.
Solution Here only tan x occurs, so we use
tan2 x = sec2 x − 1 to rewrite a tan2 x factor in
terms
of sec2 x:R
R
tan3 x dx = tan x tan2 x dx
R
Example Find tan3 x dx.
Solution Here only tan x occurs, so we use
tan2 x = sec2 x − 1 to rewrite a tan2 x factor in
terms
of sec2 x:R
R
3
2
R tan x dx 2= tan x tan x dx =
tan x (sec x − 1) dx
R
Example Find tan3 x dx.
Solution Here only tan x occurs, so we use
tan2 x = sec2 x − 1 to rewrite a tan2 x factor in
terms
of sec2 x:R
R
3
2
R tan x dx 2= tan x tan x dx =
tan x (sec x − 1) dx =
R
R
tan x sec2 x dx − tan x dx
R
Example Find tan3 x dx.
Solution Here only tan x occurs, so we use
tan2 x = sec2 x − 1 to rewrite a tan2 x factor in
terms
of sec2 x:R
R
3
2
R tan x dx 2= tan x tan x dx =
tan x (sec x − 1) dx =
R
R
2
tan x sec2 x dx − tan x dx = tan2 x −ln | sec x|+C .
R
Example Find tan3 x dx.
Solution Here only tan x occurs, so we use
tan2 x = sec2 x − 1 to rewrite a tan2 x factor in
terms
of sec2 x:R
R
3
2
R tan x dx 2= tan x tan x dx =
tan x (sec x − 1) dx =
R
R
2
tan x sec2 x dx − tan x dx = tan2 x −ln | sec x|+C .
In the first integral we mentally substituted
u = tan x so that du = sec2 x dx
R
To evaluate
the
integrals
(a)
R
R sin mx cos nx dx,
(b) sin mx sin n; dx, or (c) cos mx cos nx dx,
use the corresponding identity:
R
To evaluate
the
integrals
(a)
R
R sin mx cos nx dx,
(b) sin mx sin n; dx, or (c) cos mx cos nx dx,
use the corresponding identity:
(a) sin A cos B = 12 [sin(A − B) + sin(A + B)]
(b) sin A sin B = 12 [cos(A − B) + cos(A + B)]
(c) cos A cos B = 12 [cos(A − B) + sin(A + B)]
R
To evaluate
the
integrals
(a)
R
R sin mx cos nx dx,
(b) sin mx sin n; dx, or (c) cos mx cos nx dx,
use the corresponding identity:
(a) sin A cos B = 12 [sin(A − B) + sin(A + B)]
(b) sin A sin B = 12 [cos(A − B) + cos(A + B)]
(c) cos A cos B = 12 [cos(A − B) + sin(A + B)]
R
Example Evaluation sin 4x cos 5x dx.
R
To evaluate
the
integrals
(a)
R
R sin mx cos nx dx,
(b) sin mx sin n; dx, or (c) cos mx cos nx dx,
use the corresponding identity:
(a) sin A cos B = 12 [sin(A − B) + sin(A + B)]
(b) sin A sin B = 12 [cos(A − B) + cos(A + B)]
(c) cos A cos B = 12 [cos(A − B) + sin(A + B)]
R
Example Evaluation sin 4x cos 5x dx.
RSolution
R
sin 4x cos 5x dx = 12 [sin(−x) + sin 9x] dx
R
To evaluate
the
integrals
(a)
R
R sin mx cos nx dx,
(b) sin mx sin n; dx, or (c) cos mx cos nx dx,
use the corresponding identity:
(a) sin A cos B = 12 [sin(A − B) + sin(A + B)]
(b) sin A sin B = 12 [cos(A − B) + cos(A + B)]
(c) cos A cos B = 12 [cos(A − B) + sin(A + B)]
R
Example Evaluation sin 4x cos 5x dx.
RSolution
R
sinR4x cos 5x dx = 12 [sin(−x) + sin 9x] dx
= 21 (− sin x + sin 9x) dx
R
To evaluate
the
integrals
(a)
R
R sin mx cos nx dx,
(b) sin mx sin n; dx, or (c) cos mx cos nx dx,
use the corresponding identity:
(a) sin A cos B = 12 [sin(A − B) + sin(A + B)]
(b) sin A sin B = 12 [cos(A − B) + cos(A + B)]
(c) cos A cos B = 12 [cos(A − B) + sin(A + B)]
R
Example Evaluation sin 4x cos 5x dx.
RSolution
R
sinR4x cos 5x dx = 12 [sin(−x) + sin 9x] dx
= 21 (− sin x + sin 9x) dx = 12 (cos x − 19 cos 9x) + C .
Table of Trigonometric Substitution
Expression
√
2
2
√a − x
2
2
√a + x
x 2 − a2
Substitution
x = a sin θ, − π2 ≤ θ ≤ π2
x = a tan θ, − π2 < θ < π2
x = a sec θ, 0 ≤ θ < π2
or π ≤ θ < 3π
2
Identity
1 − sin2 θ = cos2 θ
1 + tan2 θ = sec2 θ
sec2 θ − 1 = tan2 θ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
= 1.
Find the area enclosed by the ellipse
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
= 1.
Find the area enclosed by the ellipse
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
a2 − x 2 = a cos θ.
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
a2 − x 2 dx = a cos θ · a cos θ dθ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos θ · a cos θ dθ
= a2 cos2 θ dθ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
Ra√
A = 4b
a2 − x 2
a 0
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
π
Ra√
A = 4b
a2 − x 2 = 2ab θ + 12 sin 2θ 0
a 0
√
a2 − x 2 , x = a sin θ, 1 − sin2 θ = cos2 θ
Example
x2
a
+
y2
b
Find the area enclosed by the ellipse
= 1.
Solution
√ Solving for y givesR a √
y = ba a2 − x 2 and A = 4b
a2 − x 2 dx
a 0
Substitute
x = a sin θ, dx = a cos θ dθ and use
√
2 − x 2 = a cos θ.
a
R√
R
aR2 − x 2 dx = a cos
R θ · a cos θ dθ
= a2 cos2 θ dθ = a2 12 (1 + cos 2θ) dθ = 21 a2 (θ + 21 sin 2θ).
π
Ra√
A = 4b
a2 − x 2 = 2ab θ + 12 sin 2θ 0 = πab.
a 0