Question

How do you find the integral of `int cot^4 (3x)dx`?

Answer 1

`intcot^4(3x)dx=1/9[9x+3cot(3x)-cot^3(3x)]+c`

Explanation:

Here ,

`I=intcot^4(3x)dx`

Subst. `color(blue)(3x=u=>3dx=du=>dx=1/3du`

`I=1/3intcot^4udu`

`I=1/3intcot^2u*cot^2udu`

`I=1/3intcot^2u(csc^2u-1)duto[becausecolor(red)(csc^2theta-cot^2theta=1)]`

`I=1/3intcot^2ucsc^2udu-1/3intcot^2udu`

`I=-1/3intcot^2u(-csc^2u)du-1/3intcot^2udu`

`I=-1/3int(cotu)^2d(cotu)-1/3int(csc^2u-1)du`

`I=-1/3(cotu)^3/3-1/3{-cotu-u}+c`

`I=-1/9cot^3u+1/3cotu+1/3u+c`

`I=1/9[3u+3cotu-cot^3u]+c`

Subst. back `color(blue)(u=3x`

`I=1/9[9x+3cot(3x)-cot^3(3x)]+c`