Question

Evaluate:

`int_0^(π/2) sinx/(1 + cosx)^3 dx`

Answer 1

`int_0^(π/2) sinx/(1 + cosx)^3 dx`

Put 1 + cos x = t

∴ – sin x dx = dt

∴ sin x dx = – dt

When x = 0, t = 2

When x = `π/2`, t = 1

∴ I = `int_2^1 (-dt)/t^3`

= `int_1^2 t^-3 dt`

= `[t^-2/-2]_1^2`

= `[-1/(2t^2)]_1^2`

= `-1/8 - (-1/2)`

= `-1/8 + 1/2`

= `3/8`