Evaluate : `int_0^(pi/2)(sin^2x)/(s in x+cos x)dx`

Explanation

`"Let I"=int_(0)^(pi//2) (sin^(2)x)/(sin x+ cos x)dx ......(1)`
`rArr I=int_(0)^(pi//2) (sin^(2).((pi)/(2)-x))/(sin((pi)/(2)-x)+cos ((pi)/(2) -x))dx`
[From Property (4)]
`=int_(0)^(pi//2) (cos^(2))/(cosxx +sin x)dx " ".....(2)`
Adding eqs. (1) and(2)
`2I= int_(0)^(pi//2)(sin^(2)x +cos^(2) x)/(cos x+ sin x)dx`
`=int_(0)^(pi//2) (1)/(cos x+sin x) dx`
` =int_(0)^(pi//2) (1)/(sqrt(2)((1)/(2) cos x + (1)/(sqrt(2)) sin x))dx`
`=(1)/(sqrt(2))int_(0)^(pi//2) (1)/(cos.(pi)/(4) cos x + sin .(pi)/(4) sin x)dx`
` =(1)/(sqrt(2)) int_(0)^(pi//2) (1)/(cos(x-(pi)/(4)))dx`
` =(1)/(sqrt(2)) int_(0)^(pi//2) sec(x-(pi)/(4))dx`
`=(1)/(sqrt(2))[log |sec(x-(pi)/(4)) +tan (x-(pi)/(4))|]_(0)^(pi//2)`
`=(1)/(sqrt(2))[log sec.(pi)/(4)+tan.(pi)/(4)|-log |sec(-(pi)/(4)) +tan(-(pi)/(4))|]`
`=(1)/(sqrt(2))[log (sqrt(2)+1)- log (sqrt(2)-1)]`
`=(1)/(sqrt(2)) log |(2+1)/(sqrt(2)-1)|`
`rArr I=(1)/(2sqrt(2)) log (sqrt(2)+1)^(2)`
` =(2)/(2sqrt(2))log (sqrt(2)+1)`
`=(1)/(sqrt(2))log (sqrt(2)+1).`