`G i v e nint_0^(pi/2)(dx)/(1+sinx+cosx)=log2.` Then the value of integral `int_0^(pi/2)(sinx)/(1+sinx+cosx)dx` is equal to (a) `1/2log2` (b) `pi/2-log2` (c) `pi/4-1/2log2` (d) `pi/2+log2`

Options

A. `1/2A`
B. `(pi)/2-A`
C. `(pi)/4-1/2A`
D. `(pi)/2+A`

Answer

C

Explanation

`I=int_(0)^(pi//2) (sinxdx)/(1+sinx+cosx)=int_(0)^(pi//2)(cos xdx)/(1+sinx+cosx)`
or `2I=int_(0)^(pi//2)(sinx+cosx+1-1)/(sinx+cosx+1)dx`
`2I=(pi)/2A`
or `I=(pi)/4-1/2A`