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.