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प्रश्न
Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`
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उत्तर
Let I = `int_0^pi x*sinx*cos^2x* "d"x` ......(i)
∴ I = `int_0^pi (pi - x)*sin(pi - x)*[cos(pi - x)]^2 "d"x` ......`[∵ int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x) "d"x]`
∴ I = `int_0^pi (pi - x)*sinx( - cos x)^2 "d"x`
∴ I = `int_0^pi (pi - x)* sinx cos^2x "d"x` .....(ii)
Adding (i) and (ii), we get
2I = `int_0^pi x* sinx * cos^2x "d"x + int_0^pi (pi - x) * sinx cos^2x "d"x`
= `int_0pi (x + pi - x)* sinx cos^2x "d"x`
∴ 2I = `pi int_0^pi sinx cos^2x "d"x`
Put cos x = t
∴ − sin x dx = dt
∴ sin x dx = − dt
When x = 0, t = 1 and when x = π, t = −1
∴ 2I = `pi int_1^(-1) "t"^2 (- "dt")`
∴ I = `pi/2 int_(-1)^1 "t"^2 "dt"` .......`[∵ int_"a"^"b" "f"(x) "d"x = -int_"b"^"a" "f"(x) "d"x]`
= `pi/2 xx 2 int_0^1 "t"^2 "dt"` ......[∵ t2 is an even function]
= `pi ["t"^2/3]_0^1`
= `pi/3(1^3 - 0)`
∴ I = `pi/3`
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