Strategy for using integration by parts Recall the integration by parts formula: u dv = uv - v du .
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Strategy for using integration by parts
Recall the integration by parts formula:∫u dv = uv −
∫v du.
To apply this formula we must choose dv so that wecan integrate it! Frequently, we choose u so that thederivative of u is simpler than u. If both propertieshold, then you have made the correct choice.
Strategy for using integration by parts
Recall the integration by parts formula:∫u dv = uv −
∫v du.
To apply this formula we must choose dv so that wecan integrate it!
Frequently, we choose u so that thederivative of u is simpler than u. If both propertieshold, then you have made the correct choice.
Strategy for using integration by parts
Recall the integration by parts formula:∫u dv = uv −
∫v du.
To apply this formula we must choose dv so that wecan integrate it! Frequently, we choose u so that thederivative of u is simpler than u.
If both propertieshold, then you have made the correct choice.
Strategy for using integration by parts
Recall the integration by parts formula:∫u dv = uv −
∫v du.
To apply this formula we must choose dv so that wecan integrate it! Frequently, we choose u so that thederivative of u is simpler than u. If both propertieshold, then you have made the correct choice.
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx :
Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt :
Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx :
Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx :
u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx :
u = x2 and dv = sin 2x dx
Examples using strategy:∫
u dv = uv −∫
v du
1
∫xex dx : Choose u = x and dv = ex dx
2
∫t2et dt : Choose u = t2 and dv = et dt
3
∫ln x dx : Choose u = ln x and dv = dx
4
∫x sin x dx : u = x and dv = sin x dx
5
∫x2 sin 2x dx : u = x2 and dv = sin 2x dx
∫u dv = uv −
∫v du
Example Find∫
xex dx .
Solution Let
u = x dv = ex dx.
Thendu = dx v = ex.
Integrating by parts gives∫xex dx = xex −
∫ex dx = xex − ex + C .
∫u dv = uv −
∫v du
Example Find∫
xex dx .Solution Let
u = x dv = ex dx.
Thendu = dx v = ex.
Integrating by parts gives∫xex dx = xex −
∫ex dx = xex − ex + C .
∫u dv = uv −
∫v du
Example Find∫
xex dx .Solution Let
u = x dv = ex dx.
Thendu = dx v = ex.
Integrating by parts gives∫xex dx = xex −
∫ex dx = xex − ex + C .
∫u dv = uv −
∫v du
Example Find∫
xex dx .Solution Let
u = x dv = ex dx.
Thendu = dx v = ex.
Integrating by parts gives∫xex dx = xex −
∫ex dx
= xex − ex + C .
∫u dv = uv −
∫v du
Example Find∫
xex dx .Solution Let
u = x dv = ex dx.
Thendu = dx v = ex.
Integrating by parts gives∫xex dx = xex −
∫ex dx = xex − ex + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ln x dx .
Solution Let
u = ln x dv = dx.
Then
du =1
xdx v = x.
Integrating by parts, we get∫ln x dx = x ln x −
∫xdx
x
= x ln x −∫
dx = x ln x − x + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ln x dx .Solution Let
u = ln x dv = dx.
Then
du =1
xdx v = x.
Integrating by parts, we get∫ln x dx = x ln x −
∫xdx
x
= x ln x −∫
dx = x ln x − x + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ln x dx .Solution Let
u = ln x dv = dx.
Then
du =1
xdx v = x.
Integrating by parts, we get∫ln x dx = x ln x −
∫xdx
x
= x ln x −∫
dx = x ln x − x + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ln x dx .Solution Let
u = ln x dv = dx.
Then
du =1
xdx v = x.
Integrating by parts, we get∫ln x dx = x ln x −
∫xdx
x
= x ln x −∫
dx = x ln x − x + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ln x dx .Solution Let
u = ln x dv = dx.
Then
du =1
xdx v = x.
Integrating by parts, we get∫ln x dx = x ln x −
∫xdx
x
= x ln x −∫
dx = x ln x − x + C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.
Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdt
Then du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.
Integration by parts gives∫
t2et dt = t2et − 2∫
tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time,
this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C .
Putting this inEquation (1), we get∫
t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get
∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Find∫
t2et dt.Solution Let u = t2 dv = etdtThen du = 2tdt v = et.Integration by parts gives
∫t2et dt = t2et − 2
∫tet dt (1)
Using integration by parts a second time, this time with
u = t dv = et dt.
Thendu = dt, v = et,
and∫tet dt = tet −
∫et dt = tet − et + C . Putting this in
Equation (1), we get∫t2et dt = t2et − 2
∫tet dt = t2et − 2(tet − et + C )
= t2et − 2tet + 2et + C1 whereC1 = −2C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .
Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice.
We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x,
so integration by parts gives∫ex sin x dx = −ex cos x +
∫ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x,
and∫ex cos x dx = ex sin x −
∫ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2)
and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx .
This canbe regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral.
Adding∫
ex sin x dx to both sides, we obtain2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .
Dividing by 2 and adding the constant of integration, we get∫ex sin x dx = 1
2ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get
∫ex sin x dx = 1
2ex(sin x − cos x) + C .
∫u dv = uv −
∫v du
Example Evaluate∫
ex sin x dx .Solution Solving this integral involves integrating by partstwice. We try choosing u = ex and dv = sin x dx. Thendu = ex dx and v = − cos x, so integration by parts gives∫
ex sin x dx = −ex cos x +∫
ex cos x dx . (2)
Next we use u = ex and dv = cos x dx. Then du = ex dx,v = sin x, and∫
ex cos x dx = ex sin x −∫
ex sin x dx . (3)
We put Equation (3) into Equation (2) and we get∫ex sin x dx = −ex cos x + ex sin x −
∫ex sin x dx . This can
be regarded as an equation to be solved for the unknownintegral. Adding
∫ex sin x dx to both sides, we obtain
2∫
ex sin x dx = −e cos x + ex sin x .Dividing by 2 and adding the constant of integration, we get∫
ex sin x dx = 12ex(sin x − cos x) + C .
Trigonometric integrals
Trigonometric integrals are integrals of functionsf (x) that can be expressed as a product offunctions from trigonometry.
For example;
1 f (x) = cos3 x
2 f (x) = sin5 x cos2 x
3 f (x) = sin2 x .
Integrating such functions involve several techniquesand strategies which we will describe today.
Trigonometric integrals
Trigonometric integrals are integrals of functionsf (x) that can be expressed as a product offunctions from trigonometry. For example;
1 f (x) = cos3 x
2 f (x) = sin5 x cos2 x
3 f (x) = sin2 x .
Integrating such functions involve several techniquesand strategies which we will describe today.
Trigonometric integrals
Trigonometric integrals are integrals of functionsf (x) that can be expressed as a product offunctions from trigonometry. For example;
1 f (x) = cos3 x
2 f (x) = sin5 x cos2 x
3 f (x) = sin2 x .
Integrating such functions involve several techniquesand strategies which we will describe today.
Trigonometric integrals
Trigonometric integrals are integrals of functionsf (x) that can be expressed as a product offunctions from trigonometry. For example;
1 f (x) = cos3 x
2 f (x) = sin5 x cos2 x
3 f (x) = sin2 x .
Integrating such functions involve several techniquesand strategies which we will describe today.
Trigonometric integrals
Trigonometric integrals are integrals of functionsf (x) that can be expressed as a product offunctions from trigonometry. For example;
1 f (x) = cos3 x
2 f (x) = sin5 x cos2 x
3 f (x) = sin2 x .
Integrating such functions involve several techniquesand strategies which we will describe today.
Aside from the most basic relations such as
tan x =sin(θ)
cos(θ)and sec x =
1
cos(θ), you should
know the following trig identities:
cos2(θ) + sin2(θ) = 1.
sec2(θ)− tan2(θ) = 1.
cos2(θ) =1 + cos(2θ)
2
sin2(θ) =1− cos(2θ)
2sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2(θ)−sin2(θ) = 2 cos2(θ)−1 = 1−2 sin2(θ)
Aside from the most basic relations such as
tan x =sin(θ)
cos(θ)and sec x =
1
cos(θ), you should
know the following trig identities:
cos2(θ) + sin2(θ) = 1.
sec2(θ)− tan2(θ) = 1.
cos2(θ) =1 + cos(2θ)
2
sin2(θ) =1− cos(2θ)
2sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2(θ)−sin2(θ) = 2 cos2(θ)−1 = 1−2 sin2(θ)
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:∫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∫(1− u2) du = u − 1
3u3 + C
= sin x − 1
3sin3 x + C .
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:∫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∫(1− u2) du = u − 1
3u3 + C
= sin x − 1
3sin3 x + C .
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:∫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∫(1− u2) du = u − 1
3u3 + C
= sin x − 1
3sin3 x + C .
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:∫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∫(1− u2) du = u − 1
3u3 + C
= sin x − 1
3sin3 x + C .
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:∫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∫(1− u2) du = u − 1
3u3 + C
= sin x − 1
3sin3 x + C .
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:∫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∫(1− u2) du = u − 1
3u3 + C
= sin x − 1
3sin3 x + C .
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:∫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∫(1− u2) du = u − 1
3u3 + C
= sin x − 1
3sin3 x + C .
Example Find∫ π
0sin2 x dx .
Solution Here we use the double angle formula:
sin2 x dx = 12(1− cos 2x).
∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x) dx =1
2
(x − 1
2sin 2x
)∣∣∣∣π0
=1
2(π − 1
2sin 2π)− 1
2(0− 1
2sin 0) =
1
2π.
Here we mentally made the substitution u = 2x whenintegrating cos 2x .
Example Find∫ π
0sin2 x dx .
Solution Here we use the double angle formula:
sin2 x dx = 12(1− cos 2x).∫ π
0
sin2 x dx
=1
2
∫ π
0
(1− cos 2x) dx =1
2
(x − 1
2sin 2x
)∣∣∣∣π0
=1
2(π − 1
2sin 2π)− 1
2(0− 1
2sin 0) =
1
2π.
Here we mentally made the substitution u = 2x whenintegrating cos 2x .
Example Find∫ π
0sin2 x dx .
Solution Here we use the double angle formula:
sin2 x dx = 12(1− cos 2x).∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x) dx
=1
2
(x − 1
2sin 2x
)∣∣∣∣π0
=1
2(π − 1
2sin 2π)− 1
2(0− 1
2sin 0) =
1
2π.
Here we mentally made the substitution u = 2x whenintegrating cos 2x .
Example Find∫ π
0sin2 x dx .
Solution Here we use the double angle formula:
sin2 x dx = 12(1− cos 2x).∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x) dx =1
2
(x − 1
2sin 2x
)∣∣∣∣π0
=1
2(π − 1
2sin 2π)− 1
2(0− 1
2sin 0) =
1
2π.
Here we mentally made the substitution u = 2x whenintegrating cos 2x .
Example Find∫ π
0sin2 x dx .
Solution Here we use the double angle formula:
sin2 x dx = 12(1− cos 2x).∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x) dx =1
2
(x − 1
2sin 2x
)∣∣∣∣π0
=1
2(π − 1
2sin 2π)− 1
2(0− 1
2sin 0)
=1
2π.
Here we mentally made the substitution u = 2x whenintegrating cos 2x .
Example Find∫ π
0sin2 x dx .
Solution Here we use the double angle formula:
sin2 x dx = 12(1− cos 2x).∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x) dx =1
2
(x − 1
2sin 2x
)∣∣∣∣π0
=1
2(π − 1
2sin 2π)− 1
2(0− 1
2sin 0) =
1
2π.
Here we mentally made the substitution u = 2x whenintegrating cos 2x .
Strategy for Evaluating∫
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 toexpress the remaining factors in terms of sine:
∫sinm x cos2k+1 x dx =
∫sinm x(cos2 x)k cos x dx
=
∫sinm x(1− sin2 x)k cos x dx
Then substitute u = sin x.
Strategy for Evaluating∫
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 toexpress the remaining factors in terms of sine:∫
sinm x cos2k+1 x dx =
∫sinm x(cos2 x)k cos x dx
=
∫sinm x(1− sin2 x)k cos x dx
Then substitute u = sin x.
Strategy for Evaluating∫
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 toexpress the remaining factors in terms of sine:∫
sinm x cos2k+1 x dx =
∫sinm x(cos2 x)k cos x dx
=
∫sinm x(1− sin2 x)k cos x dx
Then substitute u = sin x.
Strategy for Evaluating∫
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 toexpress the remaining factors in terms of sine:∫
sinm x cos2k+1 x dx =
∫sinm x(cos2 x)k cos x dx
=
∫sinm x(1− sin2 x)k cos x dx
Then substitute u = sin x.
Strategy for Evaluating∫
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 toexpress the remaining factors in terms of cosine:
∫sin2k+1 x cosn x dx =
∫(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∫
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 toexpress the remaining factors in terms of cosine:
∫sin2k+1 x cosn x dx =
∫(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∫
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 toexpress the remaining factors in terms of cosine:
∫sin2k+1 x cosn x dx =
∫(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∫
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 toexpress the remaining factors in terms of cosine:
∫sin2k+1 x cosn x dx =
∫(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∫
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 = 1
2(1 + cos 2x)
It is sometimes helpful to use the identity
sin x cos x = 12 sin 2x
Strategy for Evaluating∫
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 = 1
2(1 + cos 2x)
It is sometimes helpful to use the identity
sin x cos x = 12 sin 2x
Strategy for Evaluating∫
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 = 1
2(1 + cos 2x)
It is sometimes helpful to use the identity
sin x cos x = 12 sin 2x
Strategy for Evaluating∫
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 = 1
2(1 + cos 2x)
It is sometimes helpful to use the identity
sin x cos x = 12 sin 2x
Strategy for Evaluating∫
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 xto express the remaining factors in terms of tan x :
∫tanm x sec2k x dx =
∫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∫
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 xto express the remaining factors in terms of tan x :
∫tanm x sec2k x dx =
∫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∫
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 xto express the remaining factors in terms of tan x :
∫tanm x sec2k x dx =
∫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∫
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 xto express the remaining factors in terms of tan x :
∫tanm x sec2k x dx =
∫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∫
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 remainingfactors in terms of sec x :
∫tan2k+1 x secn x dx =
∫(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.
Strategy for Evaluating∫
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 remainingfactors in terms of sec x :
∫tan2k+1 x secn x dx =
∫(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.
Strategy for Evaluating∫
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 remainingfactors in terms of sec x :
∫tan2k+1 x secn x dx =
∫(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.
Strategy for Evaluating∫
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 remainingfactors in terms of sec x :
∫tan2k+1 x secn x dx =
∫(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 classusing integration by parts.
∫tan x dx = ln | sec x |+ C .
The next formula can be checked by differentiatingthe right hand side.∫
sec x dx = ln | sec x + tan x |+ C .
Also, don’t forget that ddx tan x = sec2 x and
ddx sec x = sec x tan x .
Two other useful formulas
Recall that we proved the following formula is classusing integration by parts.∫
tan x dx = ln | sec x |+ C .
The next formula can be checked by differentiatingthe right hand side.∫
sec x dx = ln | sec x + tan x |+ C .
Also, don’t forget that ddx tan x = sec2 x and
ddx sec x = sec x tan x .
Two other useful formulas
Recall that we proved the following formula is classusing integration by parts.∫
tan x dx = ln | sec x |+ C .
The next formula can be checked by differentiatingthe right hand side.
∫sec x dx = ln | sec x + tan x |+ C .
Also, don’t forget that ddx tan x = sec2 x and
ddx sec x = sec x tan x .
Two other useful formulas
Recall that we proved the following formula is classusing integration by parts.∫
tan x dx = ln | sec x |+ C .
The next formula can be checked by differentiatingthe right hand side.∫
sec x dx = ln | sec x + tan x |+ C .
Also, don’t forget that ddx tan x = sec2 x and
ddx sec x = sec x tan x .
Two other useful formulas
Recall that we proved the following formula is classusing integration by parts.∫
tan x dx = ln | sec x |+ C .
The next formula can be checked by differentiatingthe right hand side.∫
sec x dx = ln | sec x + tan x |+ C .
Also, don’t forget that ddx tan x = sec2 x and
ddx sec x = sec x tan x .
Example Find∫
tan3 x dx .
Solution Here only tan x occurs, so we usetan2 x = sec2 x− 1 to rewrite a tan2 x factor interms of sec2 x :∫
tan3 x dx =∫
tan x tan2 x dx =∫tan x (sec2 x − 1) dx =∫tan x sec2 x dx−
∫tan x dx = tan2 x
2 −ln | sec x |+C .In the first integral we mentally substitutedu = tan x so that du = sec2 x dx
Example Find∫
tan3 x dx .Solution Here only tan x occurs, so we usetan2 x = sec2 x− 1 to rewrite a tan2 x factor interms of sec2 x :
∫tan3 x dx =
∫tan x tan2 x dx =∫
tan x (sec2 x − 1) dx =∫tan x sec2 x dx−
∫tan x dx = tan2 x
2 −ln | sec x |+C .In the first integral we mentally substitutedu = tan x so that du = sec2 x dx
Example Find∫
tan3 x dx .Solution Here only tan x occurs, so we usetan2 x = sec2 x− 1 to rewrite a tan2 x factor interms of sec2 x :∫
tan3 x dx =∫
tan x tan2 x dx
=∫tan x (sec2 x − 1) dx =∫tan x sec2 x dx−
∫tan x dx = tan2 x
2 −ln | sec x |+C .In the first integral we mentally substitutedu = tan x so that du = sec2 x dx
Example Find∫
tan3 x dx .Solution Here only tan x occurs, so we usetan2 x = sec2 x− 1 to rewrite a tan2 x factor interms of sec2 x :∫
tan3 x dx =∫
tan x tan2 x dx =∫tan x (sec2 x − 1) dx
=∫tan x sec2 x dx−
∫tan x dx = tan2 x
2 −ln | sec x |+C .In the first integral we mentally substitutedu = tan x so that du = sec2 x dx
Example Find∫
tan3 x dx .Solution Here only tan x occurs, so we usetan2 x = sec2 x− 1 to rewrite a tan2 x factor interms of sec2 x :∫
tan3 x dx =∫
tan x tan2 x dx =∫tan x (sec2 x − 1) dx =∫tan x sec2 x dx−
∫tan x dx
= tan2 x2 −ln | sec x |+C .
In the first integral we mentally substitutedu = tan x so that du = sec2 x dx
Example Find∫
tan3 x dx .Solution Here only tan x occurs, so we usetan2 x = sec2 x− 1 to rewrite a tan2 x factor interms of sec2 x :∫
tan3 x dx =∫
tan x tan2 x dx =∫tan x (sec2 x − 1) dx =∫tan x sec2 x dx−
∫tan x dx = tan2 x
2 −ln | sec x |+C .
In the first integral we mentally substitutedu = tan x so that du = sec2 x dx
Example Find∫
tan3 x dx .Solution Here only tan x occurs, so we usetan2 x = sec2 x− 1 to rewrite a tan2 x factor interms of sec2 x :∫
tan3 x dx =∫
tan x tan2 x dx =∫tan x (sec2 x − 1) dx =∫tan x sec2 x dx−
∫tan x dx = tan2 x
2 −ln | sec x |+C .In the first integral we mentally substitutedu = tan x so that du = sec2 x dx
To evaluate the integrals (a)∫
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)]
Example Evaluation∫
sin 4x cos 5x dx .Solution∫
sin 4x cos 5x dx =∫
12[sin(−x) + sin 9x ] dx
= 12
∫(− sin x +sin 9x) dx = 1
2(cos x− 19 cos 9x)+C .
To evaluate the integrals (a)∫
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)]
Example Evaluation∫
sin 4x cos 5x dx .Solution∫
sin 4x cos 5x dx =∫
12[sin(−x) + sin 9x ] dx
= 12
∫(− sin x +sin 9x) dx = 1
2(cos x− 19 cos 9x)+C .
To evaluate the integrals (a)∫
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)]
Example Evaluation∫
sin 4x cos 5x dx .
Solution∫sin 4x cos 5x dx =
∫12[sin(−x) + sin 9x ] dx
= 12
∫(− sin x +sin 9x) dx = 1
2(cos x− 19 cos 9x)+C .
To evaluate the integrals (a)∫
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)]
Example Evaluation∫
sin 4x cos 5x dx .Solution∫
sin 4x cos 5x dx =∫
12[sin(−x) + sin 9x ] dx
= 12
∫(− sin x +sin 9x) dx = 1
2(cos x− 19 cos 9x)+C .
To evaluate the integrals (a)∫
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)]
Example Evaluation∫
sin 4x cos 5x dx .Solution∫
sin 4x cos 5x dx =∫
12[sin(−x) + sin 9x ] dx
= 12
∫(− sin x +sin 9x) dx
= 12(cos x− 1
9 cos 9x)+C .
To evaluate the integrals (a)∫
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)]
Example Evaluation∫
sin 4x cos 5x dx .Solution∫
sin 4x cos 5x dx =∫
12[sin(−x) + sin 9x ] dx
= 12
∫(− sin x +sin 9x) dx = 1
2(cos x− 19 cos 9x)+C .
Table of Trigonometric Substitution
Expression Substitution Identity√a2 − x2 x = a sin θ, −π
2 ≤ θ ≤ π2 1− sin2 θ = cos2 θ√
a2 + x2 x = a tan θ, −π2 < θ < π
2 1 + tan2 θ = sec2 θ√x2 − a2 x = a sec θ, 0 ≤ θ < π
2 sec2 θ − 1 = tan2 θ
or π ≤ θ < 3π2
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.
∫ √a2 − x2 dx =
∫a cos θ · a cos θ dθ
= a2∫
cos2 θ dθ = a2∫
12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ
= a2∫
cos2 θ dθ = a2∫
12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ
= a2∫
12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ
= 12a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2
= 2ab[θ + 1
2sin 2θ
]π
0= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0
= πab.
√a2 − x2, x = a sin θ, 1− sin2 θ = cos2 θ
Example Find the area enclosed by the ellipsex2
a+ y2
b= 1.
Solution Solving for y givesy = b
a
√a2 − x2 and A = 4b
a
∫ a
0
√a2 − x2 dx
Substitute x = a sin θ, dx = a cos θ dθ and use√a2 − x2 = a cos θ.∫ √
a2 − x2 dx =∫
a cos θ · a cos θ dθ= a2
∫cos2 θ dθ = a2
∫12(1 + cos 2θ) dθ = 1
2a2(θ + 1
2sin 2θ).
A = 4ba
∫ a
0
√a2 − x2 = 2ab
[θ + 1
2sin 2θ
]π
0= πab.
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