Answer 1

`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`