Let ∫₀^(π/2) [a sinx + b cosx] / [(a+b)(sinx + cosx)] dx = k. What is the value of k?
A.π/4✓ Correct
B.π/2
C.π
D.2π
Explanation
Let I = ∫₀^(π/2) (a sinx + b cosx)/[(a+b)(sinx+cosx)] dx. Apply property: replace x with π/2 - x: I = ∫₀^(π/2) (a cosx + b sinx)/[(a+b)(cosx+sinx)] dx. Adding both: 2I = ∫₀^(π/2) [(a+b)(sinx+cosx)]/[(a+b)(sinx+cosx)] dx = ∫₀^(π/2) 1 dx = π/2. So I = π/4, hence k = π/4.
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