Evaluate: `int_0^(pi/2) sin^2x/(sinx+cosx)dx`

Explanation

Let `I=int(sin^(2)xdx)/(sinx+cosx)`………………….1
`:.I=int_(0)^(pi//2)(sin^(2)(1/2pi-x)dx)/("sin"(1/2pi-x)+"cos"(1/2pi-x))`
`=int_(0)^(pi//2)(cos^(2)xdx)/(cosx+sinx)` ………..2
Now adding 1 and 2 we get
`2I=int_(0)^(pi//2)((sin^(2)x+cos^(2)x)dx)/(sinx+cosx)`
or `I=1/2int_(0)^(pi//2)(dx)/(sinx+cosx)`
`=1/(2sqrt(2))int_(0)^(pi//2)(dx)/(sin(x+1/4pi))`
`=(1/(2sqrt(2)))int_(0)^(pi//2)cosec(x+1/4)dx`
`=(1/(2sqrt(2)))[log{cosec(x+1/4pi)-cot(x+1/4pi)}]_(0)^(pi//2)`
`=(1//2sqrt(2))[log{cosec(1/2pi+1/4pi)-cot(1/2pi+1/4pi)}`
`-log{cosec(1/4pi)-cot(1/4pi)}]`
`=(1//2sqrt(2)[log{sec(1/4pi)+tan(1/4pi)}-log(sqrt(2)-1)]`
`=(1//2sqrt(2))[log(sqrt(2)+1)-log(sqrt(2)-1)]`
`=(1/(2sqrt(2)))log((sqrt(2)+1)/(sqrt(2)-1))`