Question

How do you integrate `int (tan2x+cot2x)^2` using substitution?

Answer 1

`=1/2tan 2x-1/2cot2x+C`

Explanation:

There is no need to use substitution, it makes the integral more difficult. Simple trig identities will reduce to it to something more manageable

`int((tan2x+cot2x)^2)dx`

multiply out

`int(tan^2 2x+2tan2xcot2x+cot^2 2x)dx`

`tan2x=1/(cot2x)=>tan2xcot2x=1`

`=int(tan^2 2x+2+cot^2 2x)dx`

now `1+tan^2 theta=sec^2 theta`

and `1+cot^2 theta =csc^2 theta`

so the integral becomes

`int(sec^ 2 2x +csc^2 2x)dx`

these are standard integrals

so we have

`=1/2tan 2x-1/2cot2x+C`

Answer 2

If you wanted to use substitution this is a possible solution

we have to rearrange the integral first

`int(tan 2x+cot2x)^2dx=int (tan2x+1/(tan2x))^2dx`

`=int ((tan^2 2x+1)/(tan2x))^2dx=int((tan^2 2x+1)(tan^2 2x+1))/(tan^2 2x)dx`

`=int(sec^2 2x)(tan^2 2x+1)/(tan^2 2x)dx=`

now substitute

`u=tan2x=>du=2sec^ 2xdx`

`=intcancel((sec^2 2x))(u^2+1)/u^2xx(du)/(2cancel((sec^2 2x))`

`=1/2int(1+1/u^2)dx`

`=1/2(u-1/u)+C`

`=1/2(tan2x-1/tan2x)+C`

`=1/2(tan2x-cot2x)+C`