Question

Find: `I=intdx/(sinx+sin2x)`

Find: `I=int (d theta)/(sintheta+sin2theta)`

Answer 1

`I=intdx/(sinx+sin2x)`

`=int1/(sinx+2sinxcosx)dx`

`=int1/(sinx(1+2cosx))dx`

`=intsinx/(sin^2x(1+2cosx))dx`

Let u=cosx

⇒du=−sinxdx

Also,

`sin^2x=1−cos^2x=1−u^2`

`∴ I = ∫−1/((1−u^2)(1+2u))du`

`=int 1/((1+u)(1-u)(1+2u))du`

Using partial fractions, we get

`1/((1+u)(1-u)(1+2u))=A/(1+u)+B/(1-u)+C/(1+2u)`

`=>−1=A(1−u)(1+2u)+B(1+u)(1+2u)+C(1+u)(1−u)`

`⇒−1=A(1+u−2u^2)+B(1+3u+2u^2)+C(1−u^2)`

`⇒−1=(−2A+2B−C)u^2+(A+3B)u+(A+B+C)`

Equating the respective coefficients on the LHS and the RHS, we get

−2A+2B−C=0       .....(1)

A+3B=0                 .....(2)

A+B+C=−1         .....(3)

Adding (1), (2) and (3), we get

6B=−1

⇒B=−1/6

From (2), we get

A=−3B

⇒A=1/2

From (3), we get

C=−1−A−B

⇒C=−4/3

So,

`1/((1+u)(1-u)(1+2u))=1/(2(1+u))-1/(6(1-u))-4/(3(1+2u))`

`=>I=int[1/(2(1+u))-1/(6(1-u))-4/(3(1+2u))]du`

`= 1/2log(1+u)+1/6log(1−u)−4/(xx2)log(1+2u)+C`

`= 1/2log(1+cosx)+1/6log(1−cosx)−2/3log(1+2cosx)+C`