Evaluate `int_(pi//5)^(3pi//10)(sinx)/((sinx+cosx))dx`.

Explanation

Let `I=int_(pi//5)^(3pi//10)(sinx)/(sinx+cosx)dx" …(1)" `
We use the property, `int_(a)^(b)f(x)dx=int_(a)^(b)(a+b-x)dx`.
Here `a=(pi)/(5), b=(3pi)/(10).` Hence changing x by `(pi)/(5)+(3pi)/(10)-x,` we get,
`I=int_(pi//5)^(3pi//10)(sin((pi)/(5)+(3pi)/(10)-x))/(sin((pi)/(5)+(3pi)/(10-x))+cos((pi)/(5)+(3pi)/(10-x)))`
`=int_(pi//5)^(3pi//10)(sin((pi)/(2)-x))/(sin((pi)/(2)-x)+cos((pi)/(2)-x))dx`
`int_(pi//5)^(3pi//10)(cos x)/(cos x+sinx)dx`
Adding (1) and (2), we get
`2I=int_(pi//5)^(3pi//10)(sinx)/(sinx+cosx)dx+int_(pi//5)^(3pi//10)(cos x)/(cos x+sinx)dx`
`=int_(pi//5)^(3pi//10)(sinx+cosx)/(sinx+cosx)dx" "=int_(pi//5)^(3pi//10)1 dx=[x]_(pi//5)^(3pi//10)`
`=(3pi)/(10)-(pi)/(5)=(pi)/(10)" "therefore I=(pi)/(20).`