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

2 Answers
Mar 28, 2017

#=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#

Mar 28, 2017

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#