Question

What is the integral of `sin^2(x)cos^4(x)`?

Answer 1

`1/192(12x+3sin2x-3sin4x-sin6x)+C`.

Explanation:

Let `I=intsin^2xcos^4xdx`.

We will use the following Identities to simplify the Integrand :-

`[1] :2sin^2theta=1-cos2theta, [2] : 2cos^2theta=1+cos2theta`

`[3] : 2cosCcosD=cos(C+D)+cos(C-D)`

Now, `sin^2xcos^4x=1/8(4sin^2xcos^2x)(2cos^2x)`

`=1/8(2sinxcosx)^2(1+cos2x)`

`=1/8(sin2x)^2(1+cos2x)`

`=1/8(sin^2 2x)(1+cos2x)`

`=1/16(2sin^2 2x)(1+cos2x)`

`=1/16(1-cos4x)(1+cos2x)`

`=1/16(1-cos4x+cos2x-cos4xcos2x)`

`=1/16{1-cos4x+cos2x-1/2(cos6x+cos2x)}`

`=1/32(2+cos2x-2cos4x-cos6x)`

`:. I=1/32int(2+cos2x-2cos4x-cos6x)dx`

`=1/32(2x+sin(2x)/2-(2sin(4x))/4-sin(6x)/6)`

`=1/192(12x+3sin2x-3sin4x-sin6x)+C`.

Enjoy Maths.!