Question

∫log[(a + b sin x)/(a + b cos x)]dx for x ∈ [0, π/2]is equal to....... 

(a) 0 

(b) (π/2) 

(c) (π/4) 

(d) πab

Answer 1

The correct option  (a) 0   

Explanation:

I = ^(π/2)∫₀ log[(a + bsinx)/(a + b cos x)]dx .....(1) 

= ^(π/2)∫₀ log[{a + b sin[(π/2) – x]}/{a + b cos[(π/2) – x]}]dx 

∴ I = ^(π/2)∫₀ log[(a + bcosx)/(a + b sin x)]dx  ....(2) 

(1) + (2) given, 

2I = ^(π/2)∫₀ log[(a + bsinx)/(a + bcosx)] + log[(a + bcosx)/(a + bsinx)] ∙dx 

2I = ^(π/2)∫₀ log(1)dx 

∴ 2I = 0 

I = 0.