Evaluate: `int_0^(pi/4)(sinx+cosx)/(9+16sin2x)dx`

Explanation

`I=int_(0)^(pi//4)(sinx+cosx)/(9+16 sin 2x)dx`
हम जानते हैं
`(sinx-cosx)^(2)=sin^(2)x+cos^(2)x-2sinxcosx`
`rArr(sinx-cosx)^(2)=1-sin2x`
`rArr" "sin 2x=1-(sin x-cos x)^(2)`
माना`" "sinx-cosx=t`
`rArr" "(cosx+sinx).dx=dt`
तथा `" "x=0 rArr t=0-cos 0 =-1,`
`x=(pi)/(4)rArr t=sin.(pi)/(4)-cos.(pi)/(4)=0`
`therefore" "I=int_(-1)^(0)(1)/(9+16(1-t^(2)))dx`
`=int_(-1)^(0)(1)/(9+16-16t^(2))dt`
`=int_(-1)^(0)(1)/(25-16t^(2))dt=(1)/(16)int_(-1)^(0)(1)/(((5)/(4))^(2)-t^(2))`
`=(1)/(16).(1)/(2.(5)/(4))[log|((5)/(4)+t)/((5)/(40-t))|]_(1)^(0)`
`=(1)/(40)[log|(5+4t)/(5-4t)|]_(-1)^(0)=(1)/(40)[log1-log.(1)/(9)]`
`=(1)/(40)[log1-(log1-log9)]=(1)/(40)log9`