Question

How do you find the integral of `int (cotx)^5(sinx)^4dx`?

Answer 1

`t^4/4+t^2/2-1/2*ln|t+1|-1/2ln|t-1|+C` where `t=cos(x)`

Explanation:

Note that
`cot^4(x)*sin^4(x)=cos^5(x)/sin^5(x)*sin^4(x)=cos^5(x)/sin(x)`
Substituting

`t=cos(x)`
then we get

`dt=-sin(x)dx`
and our integral will be
`-int t^5/(1-t^2)dt`
and this is
`-int( -t^3-t-1/(2*(t+1))-1/(2*(t-1)))dt`
this is
`-(t^4/4-t^2/2-1/2*ln|t+1|-1/2*ln|t-1|)+C`

Answer 2

`I=ln|sinx|-sin^2x+1/4sin^4x+c`

Explanation:

Here,

`I=int(cotx)^5(sinx)^4dx`

`=intcos^5x/sin^5x xxsin^4xdx`

`=intcos^5x/sinxdx`

`=int((cos^2x)^2cosx)/sinxdx`

`=int((1-sin^2x)^2cosx)/sinxdx`

Subst. `sinx=u=>cosxdx=du`

So,

`I=int(1-u^2)^2/udu`

`=int(1-2u^2+u^4)/udu`

`=int[1/u-2u+u^3]du`

`=ln|u|-2u^2/2+u^4/4+c`

Subst. back .`u=sinx`

`I=ln|sinx|-sin^2x+1/4sin^4x+c`