Question

How do you find the indefinite integral of `intdx/(cos(5x)-1)`?

Answer 1

`1/(5tan((5x)/2))+C`

Explanation:

`t=tan((5x)/2) => (5x)/2=arctant => dx=(2dt)/(5(1+t^2))`

We will find `cos5x` from:

`tan^2(theta/2)=(1-costheta)/(1+costheta)`

`tan^2(theta/2)=(2-1-costheta)/(1+costheta)=2/(1+costheta)-1`

`1+costheta=2/(tan^2(theta/2)+1)`

`costheta=2/(tan^2(theta/2)+1)-1`

`costheta=(2-tan^2(theta/2)-1)/(tan^2(theta/2)+1)`

`costheta=(1-tan^2(theta/2))/(tan^2(theta/2)+1)`

In our case:

`cos5x=(1-t^2)/(1+t^2)`

`I=int dx/(cos5x-1)=int ((2dt)/(5(1+t^2)))/((1-t^2)/(1+t^2)-1)`

`I=int ((2dt)/(5(1+t^2)))/((1-t^2-1-t^2)/(1+t^2))=2/5intdt/(-2t^2)=-1/5intdt/t^2`

`I=1/(5t)+C=1/(5tan((5x)/2))+C`

Answer 2

Refer to explanation

Explanation:

First set `u=5x` hence `du=5dx=>dx=1/5du`

hence the integral becomes

`int 1/5*1/(cosu-1)du=1/5*int 1/(cosu-1)du`

Now we must calculate `int 1/(cosu-1)du` hence we have that

Recall the half-angle identity:

`sin²(u/2) = (1 - cosu)/2`

hence:

`(1 - cosu) = 2sin²(u/2)=> (cosu - 1) = - 2sin²(u/2)`

The integral thus becoming:

`int 1/(- 2sin²(u/2))du =`

`=int - (1/2) 1 /(sin²(u/2)) du =`

`=int - (1/2) csc²(x/2) dx`

that is to say:

`int [- csc²(u/2)] d(u/2) =`

`int (dcot(u/2))(du)`

thus, in conclusion:

`int 1/(cosu - 1)du = cot(u/2) + c`

Hence replacing with `u=5x` we get that

`int 1/(cos5x-1)dx=1/5*cot((5x)/2)+c`