Evaluate `int_(0)^((pi)/2)(sin3x)/(sinx+cosx) dx`.

Explanation

`I=ubrace(int_(0)^((pi)/2)(sin3x-sinx)/(sinx+cosx)dx)_(I_(1))+ubrace(int_(0)^((pi)/2)(sinx)/(sinx+cosx)dx)_(I_(2))`
`I_(2)=int_(0)^((pi)/2)(sinx)/(sinx+cosx)dx`
`:.I_(2)int_(0)^((pi)/2)(cosx)/(cosx+sinx)dx`
Adding we get
`2I_(2)=(pi)/2`
`implies I_(2)=(pi)/4`
`I_(1)=int_(0)^((pi)/2)(2(cos2x)sinx)/(sinx+cosx)dx`
`=2int_(0)^((pi)/2)sinx(cosx-sinx)dx`
`=int_(0)^((pi)/2)sin2xdx-int_(0)^((pi)/2)2sin^(2) xdx`
`=[-(cos2x)/2]_(0)^((pi)/2)-int_(0)^((pi)/2)(1-cos2x)dx`
`=1-(pi)/4`
Therefore `I=1`