Question

How do you find the antiderivative of `cos^4 (2x) * sin^3 (2x)`?

Answer 1

`intcos^4(2x)*sin^3(2x)dx=cos^7(2x)/14-cos^5(2x)/10+C`

Explanation:

We have the integral `I_(color(red)1 )=intcos^4(2x)*sin^3(2x)dx`

Here, let

`(2x)=t`

`d/dx(2x)=d/dxt`

`2=dt/dx`

Therefore, `dx=dt/2`

Put this in `I_(color(red)1`

`I_(color(red)2 )=1/2intcos^4(t)*sin^3(t)dt`

`I_(color(red)2 )=1/2intcos^4(t)*sin^2(t)*sin(t)dt`

We know `(sin^2t=1-cos^2t)`

Put this in `I_(color(red)2`

`I_(color(red)2 )=1/2intcos^4(t)*(1-cos^2t)*sin(t)dt`

Here, let

`cost=v`

`d/dtcost=d/dtv`

`-sint=(dv)/dt`

Therefore,

`sin(t)dt=-dv`

Put this in `I_(color(red)2`

`I_(color(red)3 )=(-1)/2intv^4*(1-v^2)*dv`

`I_(color(red)3 )=(-1)/2int(v^4-v^6)*dv`

`I_(color(red)3 )=((-1)/2)(v^5)/5-((-1)/2)v^7/7+C`

Now put in the value of `v`.

`I_(color(red)3 )=((-1)/2)(cos^5(t))/5-((-1)/2)cos^7(t)/7+C`

Now put in the value of `t`.

`I_(color(red)3 )=(-cos^5(2x))/10+cos^7(2x)/14+C`

Therefore,

`intcos^4(2x)*sin^3(2x)dx=cos^7(2x)/14-cos^5(2x)/10+C`