Question

evaluate ∫ cos x / root[sin^2x - 2sinx - 3 ]dx

Answer 1

Given,

∫ cos x / √[sin^2x - 2sinx - 3 ]dx

Let t = sinx then, cos x dx = dt

= dt / √(t² - 2t  - 3)

= dt / √(t² - 2t + 1 -1 -3)

= dt / √((t-1)² - (2)²)

= log | (t-1) + √(t-1)^2 - (2)^2 | + C

= log | (sin x - 1) + √sin²x-2sinx-3 | + C