# Find the indefinite integral `int 2/(sec(x) -1) dx`

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### 1 Answer

The integral `int 2/(sec x - 1) dx` has to be determined.

`int 2/(sec x - 1) dx`

= `int 2/((1/cosx) - 1) dx`

= `int 2 /((1 - cosx) /cosx) dx`

= `int (2*cos x) /(1 - cosx)dx`

= `int cos x/((1 - cosx)/2)dx`

The denominator `(1-cos x)/2` can be written as `sin^2(x/2)` . The numerator can be written as `cos x = cos^2(x/2) - sin^2(x/2)`

`int cos x/((1 - cosx)/2)dx`

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

= `int cot^2(x/2) - 1 dx`

Use the relation `cot^2(x/2) = cosec^2(x/2) - 1`

= `int cosec^2(x/2) - 1 - 1 dx`

= `int cosec^2(x/2) - 2 dx`

= `int cosec^2(x/2) dx - int 2 dx`

= `-2*cot(x/2) - 2x + C`

**The integral **`int 2/(sec x - 1) dx = -2*cot(x/2) - 2x + C`