# Find: I=intdx/(sinx+sin2x) - Mathematics and Statistics

Sum

Find: I=intdx/(sinx+sin2x)

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

#### Solution

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+2BC=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=1AB

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

Concept: Methods of Integration: Integration Using Partial Fractions
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