Question

How do you evaluate `int cot^5(2x) dx`?

Answer 1

`I=1/8(4ln(sin(2x))-csc^4(2x)+4csc^2(2x))+C`

Explanation:

We want to integrate

`I=intcot^5(2x)dx`

Make a substitution `color(red)(u=2x=>du=2dx`

`I=1/2intcot^5(u)du`

`color(white)(I)=1/2int((cos^2(u))^2cos(u))/sin^5(u)du`

`color(white)(I)=1/2int((1-sin^2(u))^2cos(u))/sin^5(u)du`

Make a substitution `color(red)(s=sin(u)=>ds=cos(u)du`

`I=1/2int(1-s^2)^2/s^5ds`

`color(white)(I)=1/2int1/s+1/s^5-2 1/s^3ds`

`color(white)(I)=1/2(ln(s)-1/(4s^4)+1/s^2)+C`

`color(white)(I)=1/8(4ln(s)-1/s^4+4/s^2)+C`

Substitute back `color(red)(s=sin(u)` and `color(red)(u=2x`

`I=1/8(4ln(sin(2x))-csc^4(2x)+4csc^2(2x))+C`