Question

Evaluate: `int_0^π x/(1 + sinx)dx`.

Answer 1

`int_0^π x/(1 + sinx)dx`

Let I = `int_0^π x/(1 + sinx)dx`  ...(i)

On using property

`int_0^a f(x)dx = int_0^a f(a - x)dx`

∴ I = `int_0^π (π - x)/(1 + sin(π - x))dx`

I = `int_0^π (π - x)/(1 + sinx)dx`  ...(ii)

Adding equations (i) and (ii), we get

2I = `int_0^π π/(1 + sinx)dx`

= `πint_0^π 1/(1 + sinx) xx (1 - sinx)/(1 - sinx)dx`  ...[Multiplying and dividing by (1 – sin x)]

= `πint_0^π (1 - sinx)/(1 - sin^2x)dx = πint_0^π (1 - sinx)/(cos^2x)dx`

= `π[int_0^π 1/(cos^2x)dx - int_0^π sinx/(cos^2x)dx]`

= `π[int_0^π sec^2x  dx - int_0^π secx tanx  dx]`

= `π[[tanx]_0^π - [secx]_0^π]`

= π[0 – (– 1 – 1)]

= 2π

∴ I = `(2π)/2` = π.