I=∫0πxf(sinx)dx=∫0π(π−x)f(sinx)dx =π∫0πf(sinx)dx−1 2I=π∫0πf(sinx)dx I=2π∫0πf(sinx)dx=π∫0π/2f(sinx)dx =π∫0π/2f(cosx)dx
JEE Main 2006 — Mathematics Calculus
∫0πxf(sinx)dx is equal to
Held on 30 Apr 2006 · Verified 6 Jul 2026.
π∫0πf(cosx)dx
π∫0πf(sinx)dx
2π∫0π/2f(sinx)dx
π∫0π/2f(cosx)dx
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