Question

Evaluate : `∫_0^(pi/2) (sinx.cosx)/(1 + sin^4x)`.dx

Answer 1

Let I = `∫_0^(pi/2) (sinx.cosx)/(1 + sin^4x)`.dx

Put `sin^2x  = t`

2sinx cosx dx = dt

`sinx cosx  dx = dt/2`

When x = 0 , t = 0

`x = pi/2` , t = 1

∴ I = `∫_0^1 (dt/2)/(1 + t^2)`

= `∫_0^1 1/2[dt/(1 + t^2)]`

= `[1/2 (tan^-1 t)]_0^1`

= `1/2[tan^-1 1 - tan^-1 0]`

= `1/2[pi/4 - 0]`

`"I" = pi/8`