Question

Evaluate:

`int 1/(1 + cosα . cosx)dx`

Answer 1

Let I = `int 1/(1 + cosα . cosx)dx`

Put tan`(x/2)` = t

∴ x = 2 tan^–1t

∴ dx = `(2dt)/(1 + t^2)` and cos x = `(1 - t^2)/(1 + t^2)`

∴ I = `int 1/(1 + cosα ((1 - t^2)/(1 + t^2))).(2dt)/(1 + t^2)`

= `int (1 + t^2)/(1 + t^2 + cosα - cosα.t^2).(2dt)/(1 + t^2)`

= `2int 1/((1 + cosα) + (1 - cosα)t^2)dt`

= `2int 1/(2cos^2 (α/2) + 2sin^2(α/2).t^2)dt`

= `2/(2sin^2(α/2)) int 1/(cot^2 (α/2) + t^2)dt`

= `1/(sin^2(α/2)) xx 1/(cos(α/2)).tan^-1[t/(cot(α/2))] + c`

= `2/(2sin (α/2).cos (α/2)).tan^-1 [tan(α/2).tan(x/2)] + c`

= `2  "cosec"  α . tan^-1 [tan (α/2).tan(x/2)] + c`.