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

Explanation

Let `I=int_(0)^(pi//2)(sin^(2)x)/(sinx+cosx)dx`
Using the property, `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx`, we have
`I=int_(0)^(pi//2)(sin^(2)((pi)/(2)-x))/(sin[(pi//2)-x]+cos[(pi//2)-x])dx=int_(0)^(pi//2)(cos^(2)x)/(cosx+sinx)dx" ...(2)"`
Adding (1) and (2), we get,
`2I=int_(0)^(pi//2)(sin^(2)x)/(sinx+cosx)dx+int_(0)^(pi//2)(cos^(2)x)/(cosx+sinx)dx`
`=int_(0)^(pi//2)(sin^(2)x+cos^(2)x)/(sinx+cosx)dx=int_(0)^(pi//2)(dx)/(sinx+cosx)`
Now, `sinx+cosx=sqrt2[(sinx)((1)/(sqrt2))+(cosx)((1)/(sqrt2))]`
`=sqrt2[sinxcos.(pi)/(4)+cosx sin/(pi)/(4)]=sqrt2 sin (x+(pi)/(4))`
`therefore 2I=int_(0)^(pi//2)(dx)/(sqrt2sin (x+(pi)/(4)))=(1)/(sqrt2)int_(0)^(pi//2)" cosec "(x+(pi)/(4))dx`
`=(1)/(sqrt2)[log{"cosec "(x+(pi)/(4))-cot (x+(pi)/(4))}]_(0)^(pi//2)`
`=(1)/(sqrt2)[log("cosec"(3pi)/(4)-cot.(3pi)/(4))-log("cosec "(pi)/(4)-cot.(pi)/(4))]`
`=(1)/(sqrt2)[log(sqrt2+1)-log(sqrt2-1)]`
`[{:("as cosec "(3pi)/(4)="cosec "(pi-(pi)/(4))="cosec "(pi)/(4)=sqrt2),("and cot "(3pi)/(4)=cot(pi-(pi)/(4))=-cot.(pi)/(4)=-1):}]`
`=(1)/(sqrt2)log((sqrt2+1)/(sqrt2-1))=(1)/(sqrt2)log((sqrt2+1)/(sqrt2-1).(sqrt2+1)/(sqrt2+1))`
`=(1)/(sqrt2)log.((sqrt2+1)^(2))/(2-1)=(2)/(sqrt2)log(sqrt2+1)`
`therefore I=(1)/(sqrt2)log(sqrt2+1).`