The value of `int_(pi//4)^(3pi//4)(x)/(1+sinx)` dx .. . . . .

Answer

`pi(sqrt(2)-1)`

Explanation

Let `rArr I=int_(pi//4)^(3pi//4)(x)/(1+sinx)dx` . . . (i) `rArrI=int_(pi//4)^(3pi//4)(((pi)/(4)+(3pi)/(4)-x))/(1+sin (pi/(4)+(3pi)/(4)x))dx`
`[:' int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx]`
`=int_(pi//4)^(3pi//4)(pi-x)/(1+sin(pi-x))dx`
`=int_(pi//4)^(3pi//4)(pi)/(1+sinx)dx-int_(pi//4)^(3pi//4)(x)/(1+sinx)dx` `=int_(pi//4)^(3pi//4)(dx)/(1+sinx)-I ` [ from Eq . (I)] `=(pi)/(2)int_(pi//4)^(3pi//4)(dx)/((1+sinx))`
`= (pi)/(2)int_(pi//4)^(3pi//4)((1-sinx))/((1+sinx)(1-sinx))dx` `=(pi)/(4)int_(pi//4)^(3pi//4)((1-sinx))/(1-sinx)dx`
`=(pi)/(2)int_(pi//4)^(3pi//4)((1)/(cos^(2)x)-(sinx)/(cos^(2)x))dx`
`=(pi)/(2)int_(pi//4)^(3pi//4)(sec^(2)x-secx*tanx)dx`
`=(pi)/(2)[tanx-sec x]_(pi//4)^(3pi//4)`
`=(pi)/(2)[-1-1-(-sqrt(2)-sqrt(2))]`
`=(pi)/(2)(-2+2sqrt(2))=pi(sqrt(2)-1)`