Question

Evaluate the following : `int_0^pi x sin x cos^2x*dx`

Answer 1

Let I = `int_0^pi x sin x cos^2x*dx`

= `(1)/(2) int_0^a x(2 sin x cos x)cos x*dx`

= `(1)/(2) int_0^pi x(sin 2x cos x)*dx`

= `(1)/(4) int_0^pi x (2 sin 2x cos x)*dx`

= `(1)/(4) int_0^pi [sin(2x + x) + sin(2x - x)]*dx`

= `(1)/(4)[int_0^pi x sin 3x*dx + int_0^pi x sin x*dx]`

= `(1)/(4)["I"_1 + "I"_2]`                          ...(1)

I₁ = `int_0^pi x sin 3x*dx`

= `[x int sin 3x*dx]_0^pi - int[{d/dx (x) int sin 3x*dx}]*dx`

= `[x((- cos3x)/3)]_0^pi - int_0^pi 1((- cos 3x)/3)*dx`

= `[- (pi cos 3pi)/3 + 0] + (1)/(3) int_0^pi cos 3x*dx`

= `- pi/(3)(- 1) + 1/3 [(sin3x)/3]_0^pi`

= `pi/(3)  + (1)/(3)[0 - 0]`

= `pi/(3)`                                          ...(2)

I₂ = `int_0^pi x sinx*dx`

= `[x int sinx*dx]_0^pi - int_0^pi[{d/dx (x) int sinx*dx}]*dx`

= `[x(- cos x)]_0^pi - int_0^pi 1*(- cos x)*dx`

= `[ - pi cospi + 0] + int_0^pi cosx*dx`

= `-pi(-1) + [sin x]_0^pi`

= `pi + [sin pi - sin 0]`

= `pi + (0 - 0)`
= π                                            ...(3)
From (1), (2) and (3), we get

I = `(1)/(4)[pi/3 + pi]`

= `(1)/(4)((4pi)/3)`

= `pi/(3)`.