Evaluate the following :
`int_(0)^(pi//2)(1)/(1+sqrt(tanx))dx`

Explanation

Let `I=int_(0)^(pi//2)(dx)/(1+sqrt(tanx))`
`=int_(0)^(pi//2)(dx)/(1+sqrt((sinx)/(cosx)))`
`=int_(0)^(pi//2)(sqrt(cosx))/(sqrt(cosx)+sqrt(sinx))dx" …(1)"`
We use the property `int_(0)^(a)f(a-x)dx.`
Hence in I, we change x by `(pi//2)-x.` Then
`I=int_(0)^(pi//2)(sqrt(cos[(pi//2)-x]))/(sqrt(cos[(pi//2)-x])+sqrt(sin[(pi//2)-x]))dx`
`=int_(0)^(pi//2)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx" ...(2)"`
Adding (1) and (2). we get,
`2I=int_(0)^(pi//2)(sqrtcosx+sqrtsinx)/(sqrtcosx+sqrtsinx)dx+int_(0)^(pi//2)(sqrtsinx)/(sqrtsinx+sqrtcosx)dx`
`=int_(0)^(pi//2)(sqrtcosx+sqrtsinx)/(sqrtcosx+sqrtsinx)dx`
`=int_(0)^(pi//2)1dx=[x]_(0)^(pi//2)`
`=(pi//2)-0=pi/2" "therefore I=pi//4.`