Question

Evaluate the following:

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

Answer 1

Let `I = int_0^pi x/(1 + sin^2x) * dx`                     ...(1)

We use the property, `int_0^a f(x) * dx = int_0^a f(a - x) * dx`

Here a = π.

Hence in I, changing x to π – x, we get

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

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

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

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

∴ `2I = pi int_0^(pi) 1/(1 + sin^2x) * dx`

Dividing numerator and denominator by cos²x, we get

`2I = pi int_0^(pi) (sec^2x)/(sec^2x + tan^2x) * dx`

= `pi int_0^(pi) (sec^2x)/(1 + 2tan^2x) * dx`

Put tan x = t

∴ sec²x dx = dt

When x = π, t = tan π = 0

When x = 0, t = tan 0 = 0

∴ `2I = pi int_0^(0) dt/(1 + 2t^2) = 0`

∴ I = 0    ...`[because int_a^a f(x) * dx = 0]`