`int (d theta)/(sin theta +sin 2theta)`

Explanation

`"Let "I=int (1)/(sin theta+sin 2 theta)d theta=int (1)/(sin theta+2sin thetacos theta)d theta`
`=int (1)/(sintheta(1+2costheta))d theta=int (sin theta)/(sin^(2)theta(1+2costheta))d theta `
`=int (sin theta)/((1-cos^(2)theta)(1+2cos theta))d theta`
`=int (sin theta)/((1-costheta)(1+costheta)(1+2costheta))d theta `
Put `cos theta=t " "therefore -sin theta=dt`
`therefore sin theta=-dt`
`therefore I=int (-dt)/((1-t)(1+t)(1+2t))`
`=-int (dt)/((1-t)(1+t)(1+2t))`
`"Let "(1)/((1-t)(1+t)(1+2t))=(A)/(1-t)+(B)/(1+t)+(C)/(1+2t)`
`therefore 1=A((1+t)(1+2t)+B(1-t)+C(1-t)(1+t)`
Putting 1-t=0, i.e, t=1 we get
`1=A(2)(3)+B(0)(3)+C(0)(2)" "thereforeA=1//6`
Putting 1+t=0, i.e, t=1, we get
1=A(0)(-1)+B(2)(-1)+C(2)(0)
`therefore B=-1//2`
Putting 1+2t=0, i.e, t=-1//2, we get
`1=A(0)+B(0)+C((3)/(2))((1)/(2))therefore C=4//3`
`therefore (1)/((1-t)(1+t)(1+2t))=((1//6))/(1-t)+((-1//2))/(1+t)+((4//3))/(1+2t)`
`therefore I=-int [((1//6))/(1-t)+((-1//2))/(1+t)+((4//3))/(1+2t)]dt`
`=-(1)/(6)int (1)/(1-t)dt+(1)/(2)int (1)/(1+t)-(4)/(3)int (1)/(1+2t)dt`
`=-(1)/(6).(log|1-t)/(-1)+(1)/(2)|1+t|-(4)/(3).(log|1+2t|)/(2)+c`
`=(1)/(6)log|1-cos theta|+(1)/(2)log|1+cos theta|-(2)/(3)log|1+2cos theta|+c`