Question

`int_0^pi x sin^2x dx` = ______ 

Options

• `pi^2/2`

• `(3pi^2)/2`

• `pi^2/3`

• `pi^2/4`

Answer 1

`int_0^pi x sin^2x dx` = `underline(pi^2/4)`

Explanation:

Let I = `int_0^pi x sin^2x dx` ............(i)

∴ I = `int_0^pi (pi - x)sin^2x dx` .............(ii) `[∵ int_0^a f(x)dx = int_0^af(a - x)dx]`

Adding (i) and (ii), we get

2I = `int_0^pi pi sin^2x dx`

= `pi int_0^pi (1 - cos2x)/2 dx`

∴ 2I = `pi/2[x - (sin2x)/2]_0^pi`

= `pi/2(pi - 0)`

∴ I = `pi^2/4`