Question

Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`

Answer 1

Let `sin x = t => cos x dx = dt`

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

Using partial fraction

`2/((1-t)(1+t^2)) = A/((1-t)) + (Bt + C)/((1+t^2))`

On solving A = 1, B =1, C = 1

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

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

`= -In (1-t) + tan^(-1) t + 1/2 In (1+t^2)`

`= In sqrt(1+t^2)/(1-t) + tan^(-1) t + C`

Replacing the value of t

`= In sqrt(1+sin^2x)/(1-sinx) + tan^(-1)(sin x) + C`