Question

How do you integrate `cscxcot^4xdx`

Answer 1

`intcsc(x)*cot^4(x)dx`

Just trigonometric function : let's `t = tan(1/2x)`

`csc(x) = (1+t^2)/(2t)`

`cot(x) = (1-t^2)/(2t)`

`dx = 2/(1+t^2)dt`

So now we have :

`int(1+t^2)/(2t)*((1-t^2)/(2t))^4*2/(1+t^2)dt`

`=> int 1/t*((1-t^2)/(2t))^4dt`

Use the Binomial theorem

`=>int1/t*(1-4t^2+6t^4-4t^6+t^8)/(16t^4)dt`

`=>1/16int(1-4t^2+6t^4-4t^6+t^8)/(t^5)dt`

`=>1/16int1/t^5-4/t^3+6/t-4t+t^3dt`

`=>1/16[-1/(4t^4)+2/t^2+6ln(|t|)-2t^2+1/4t^4]+C`

And then substitute back for `t = tan(1/2x)`...