Evaluate the following :
`int_(0)^(pi//2)(root3(secx))/(root3(secx)+root3("cosec x"))dx`

Explanation

Let `I=int_(0)^(pi//2)(root3(secx))/(root3(secx)+root3("cosec x"))dx" …(1)"`
We use the property, `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx`
Hence in I, we change x by `(pi//2)-x.` Then
`I=int_(0)^(pi//2)(root3(sec[(pi//2)-x]))/(root3(sec[(pi//2)-x])+root3("cosec "[(pi//2)-x]))dx`
`=int_(0)^(pi//2)(root3("cosec x"))/(root3("cosec x")+root3("sec x"))dx" ...(2)"`
Adding (1) and (2), we get,
`2I=int_(0)^(pi//2)(root3(secx))/(root3(secx)+root3("cosec x"))dx+int_(0)^(pi//2)(root3("cosec x"))/(root3("cosec x")+root3(sec x))dx`
`=int_(0)^(pi//2)(root3(secx)+root3(" cosec x"))/(root3(secx)+root3("cosec x"))dx=int_(0)^(pi//2)1dx=[x]_(0)^(pi//2)`
`=(pi//2)-0=pi//2" "therefore I=pi//4.`