Question

How to solve ∫ (sin^4x)(cos^2x) dx ?

Answer 1

`int sin^4x*cos^2x dx=1/16*(x-sin(4x)/4-(sin^3(2x))/3)+c`

Explanation:

`intsin^4x*cos^2x dx=int(sin^2x)*(sin^2x*cos^2x)dx`

`=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx`

We have

`cos2x=1-2sin^2x`

`sin^2x=(1-cos2x)/2`

and

`sin2x=2sinx*cosx`

The integral becomes

`=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx`

`=1/8(int(1-cos2x)*sin^2(2x))dx`

`=1/8(int(sin^2(2x)-cos2x*sin^2(2x))dx`

`=1/8(int(1/2)*2sin^2(2x)dx-int1/2*2cos2x*sin^2(2x)dx)`

`=1/8(int(1/2)*(1-cos(4x))dx-int1/2*sin^2(2x)d(sin2x))`

`=1/16(int(1-cos(4x))dx-intsin^2(2x)d(sin2x))`

`d(sin2x)/dx=2cos2x`

`d(sin2x)=2cos2x*dx`

`=1/16(x-sin(4x)/4-sin^3(2x)/3)+c`