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Question
Evaluate :`int_0^(pi/2)(2^(sinx))/(2^(sinx)+2^(cosx))dx`
Solution
`Let I=int_0^(pi/2)(2^(sinx))/(2^(sinx)+2^(cosx))dx ...........(1)`
Then
`I=int_0^(pi/2)(2^(sin(pi/2-x)))/(2^(sin(pi/2-x))+2^(cos(pi/2-x)))dx [∫_0^a f(x)dx=∫_0^a f(a−x)dx]`
`I=int_0^(pi/2)(2^(cosx))/(2^(cosx)+2^(sinx))dx .........(2)`
Adding (1) and (2), we get
`2I=int_0^(pi/2)(2^(sinx))/(2^(sinx)+2^(cosx))dx+int_0^(pi/2)(2^(cosx))/(2^(cosx)+2^(sinx))dx`
`2I=int_0^(pi/2) dx`
`2I=x_0^(pi/2)`
`2I=pi/2`
`I=pi/4`
`therefore int_0^(pi/2)(2^(sinx))/(2^(sinx)+2^(cosx))dx=pi/4`
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