Question

Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`

Answer 1

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`