∫₀^(π) sin² x dx equals:
Aπ
Bπ/2
C2π
Dπ/4
Answer & Solution
Correct answer: B. π/2
sin² x = (1 - cos 2x)/2. So ∫₀^π sin² x dx = (1/2)∫₀^π (1 - cos 2x) dx = (1/2)[x - sin(2x)/2]₀^π = (1/2)[π - 0] = π/2.
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