Question

If ∫[(cosx)/√(sin²x + 2sinx + 1)]dx = Alog√(sinx + 1) + c then A = __________ 

(a) 2 

(b) 1 

(c) (1/2) 

(d) – 2

Answer 1

The correct option  (a) 2 

Explanation:

I = ∫[(cosxdx)/√(sin²x + 2sinx + 1)] 

Put sinx = t 

∴ cosxdx = dt 

∴ I = ∫[dt/√(t² + 2t + 1)] 

I = ∫[dt/(t + 1)] 

∴ I = log(1 + t) + c 

I = log(sinx + 1) + c given is 

I = Alog√(sinx + 1) + c 

i.e. I = A × (1/2)log(sinx + 1) + c 

∴ (A/2) = 1 

∴ A = 2