Question

`int 1/(cos x - sin x)dx`

Answer 1

Given I = `int 1/(cos x - sin x)dx`

We know that sin x = `(2 tan (x/2))/(1 + tan^2 (x/2)) and cos x = (1 - tan^2 (x/2))/(1 + tan^2 (x/2))`

⇒ `int 1/(-sin x + cos x)dx = int 1/(- (2 tan (x/2))/(1 + tan^2 (x/2)) + (1 - tan^2 (x/2))/(1 + tan^2 (x/2)))`

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

Replacing 1 + tan² x/2 in numerator by sec² x/2 and putting tan x/2 = t and sec² x/2 dx = 

⇒ `int (1 + tan^2 (x/2))/(-2 tan (x/2) + 1 - tan^2 (x/2))dx`

= `int (sec^2 (x/2))/(- tan^2 (x/2) - 2 tan (x/2) + 1) dx`

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

= `-2 int 1/((t + 1)^2 - (sqrt2)^2)dt`

= `2 int 1/((sqrt2)^2 - (t + 1)^2)dt`

We know that `int 1/(a^2 - x^2)dx = 1/(2a) log |(a + x)/(a - x)| + c`

= `2 int 1/((sqrt2)^2 - (t + 1)^2)dt`

= `2/(2sqrt2)log|(sqrt2 + t + 1)/(sqrt2 - t - 1)|+c`

= `1/sqrt2 log|(sqrt2 + tan (x/2)+1)/(sqrt2 - tan (x/2)-1)| + c`

= `1/sqrt2 log |(sqrt2 + tan (x/2) +1)/(sqrt2 - tan (x/2)-1)| + c`

∴ I = `int 1/(cos x - sin x)dx = 1/sqrt2 log |(sqrt2 + tan (x/2)+ 1)/(sqrt2 - tan (x/2) - 1)|+x`