\(I = \int \frac{dx}{\sin x + \sin 2x}\)
\(= \int \frac 1{\sin x + 2\sin \cos x} dx\)
\(= \int \frac 1{\sin x(1 + 2\cos x)}dx\)
\(= \int \frac{\sin x}{\sin^2 x( 1 + 2\cos x)} dx\)
Let u = cosx
⇒ du = −sinx dx
Also,
\(\sin^2x = 1 - \cos^2x \)
\(= 1 - u^2\)
\(\therefore I = \int -\frac 1{(1 - u^2)(1 + 2u)} du\)
\(= \int \frac 1{(1+ u)(1 - u)(1 + 2u)} du\)
Using partial fractions, we get
\(\frac 1{(1 + u) (1 - u)(1 + 2u)} = \frac A{1 + u} + \frac B{1 - u}+ \frac C{1 + 2u}\)
⇒ \(- 1 = A(1- u)(1 + 2u) + B(1 + u)(1 + 2u) + C(1 + u)(1 -u)\)
⇒ \(-1 = A(1 + u - 2u^2) + B(1 + 3u + 2u^2) + C(1 - u^2)\)
⇒ \(-1 = (-2A + 2B - C)u^2 + (A + 3B)u + (A + B + C)\)
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 = \(- \frac 16\)
From (2), we get
A = −3B
⇒ A = \(\frac 12\)
From (3), we get
C = −1 − A − B
⇒ C = \(-\frac 43\)
So,
\(\frac 1{(1 + u) (1 - u)(1 + 2u)} = \frac 1{2(1 + u)} - \frac 1{6(1 - u)}- \frac 4{3(1 + 2u)}\)
⇒ \(I = \int \left[\frac 1{2(1 + u)} - \frac 1{6(1 - u)} - \frac 4{3(1 + 2u)}\right]du\)
\(= \frac 12 \log(1 + u) + \frac 16 \log(1 - u) - \frac 4{3\times 2} \log(1 + 2u) + C\)
\(= \frac 12 \log(1 + \cos x) + \frac 16 \log(1 - \cos x) - \frac 23 \log(1 + 2\cos x) + C\)
