# How to integrate cosec x with respect to x?

*print*Print*list*Cite

### 1 Answer

The integral `int cosec x dx` has to be determined.

`int cosec x dx`

=> `int 1/sin x dx`

=> `int sin x/(sin^2x) dx`

=> `int sin x/(1 - cos^2x) dx`

let `y = cos x => dy = -sin x dx`

=> `int 1/(y^2 - 1) dy`

=> `(1/2)*int ((y + 1) - (y - 1))/(y^2 - 1) dy`

=> `(1/2)*int 1/(y - 1) dy - (1/2)*int 1/(y + 1) dy`

=> `(ln(y -1) - ln(y + 1))/2`

substitute y = cos x

=> `(ln(cos x - 1) - ln(cos x + 1))/2 + C`

**The integral `int cosec x dx = (ln(cos x - 1) - ln(cos x + 1))/2 + C`**