# ∫ 1 Sin X + Sin 2 X D X - Mathematics

Sum
$\int\frac{1}{\sin x + \sin 2x} dx$

#### Solution

We have,
$I = \int\frac{dx}{\sin x + \sin 2x}$
$= \int\frac{dx}{\sin x + 2 \sin x \cos x}$
$= \int\frac{dx}{\sin x \left( 1 + 2 \cos x \right)}$
$= \int\frac{\sin x dx}{\sin^2 x \left( 1 + 2 \cos x \right)}$
$= \int\frac{\sin x dx}{\left( 1 - \cos^2 x \right) \left( 1 + 2 \cos x \right)}$
$= \int\frac{\sin x dx}{\left( 1 - \cos x \right) \left( 1 + \cos x \right) \left( 1 + 2 \cos x \right)}$
$\text{Putting }\cos x = t$
$\Rightarrow - \sin x dx = dt$
$\Rightarrow \sin x dx = - dt$
$\therefore I = \int\frac{- dt}{\left( 1 - t \right) \left( 1 + t \right) \left( 1 + 2t \right)}$
$= \int\frac{dt}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)}$
$\text{Let }\frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{1 + 2t}$
$\Rightarrow \frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)} = \frac{A \left( t + 1 \right) \left( 1 + 2t \right) + B \left( t - 1 \right) \left( 1 + 2t \right) + C \left( t - 1 \right) \left( t + 1 \right)}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)}$
$\Rightarrow 1 = A \left( t + 1 \right) \left( 1 + 2t \right) + B \left( t - 1 \right) \left( 1 + 2t \right) + C \left( t - 1 \right) \left( t + 1 \right)$
$\text{Putting t + 1 = 0}$
$\Rightarrow t = - 1$
$1 = B \left( - 1 - 1 \right) \left( 1 - 2 \right)$
$\Rightarrow 1 = B \left( - 2 \right) \left( - 1 \right)$
$\Rightarrow B = \frac{1}{2}$
$\text{Putting t - 1 = 0}$
$\Rightarrow t = 1$
$1 = A \left( 1 + 1 \right) \left( 1 + 2 \right)$
$\Rightarrow 1 = A\left( 2 \right)\left( 3 \right)$
$\Rightarrow A = \frac{1}{6}$
$\text{Putting 1 + 2t = 0}$
$t = - \frac{1}{2}$
$\Rightarrow 1 = A \times 0 + B \times 0 + C \left( - \frac{1}{2} - 1 \right) \left( - \frac{1}{2} + 1 \right)$
$1 = C \left( - \frac{3}{2} \right) \left( \frac{1}{2} \right)$
$C = - \frac{4}{3}$
Then,
$I = \frac{1}{6}\int\frac{dt}{t - 1} + \frac{1}{2}\int\frac{dt}{t + 1} - \frac{4}{3}\int\frac{dt}{1 + 2t}$
$= \frac{1}{6} \log \left| t - 1 \right| + \frac{1}{2} \log \left| t + 1 \right| - \frac{4}{3} \times \frac{\log \left| 1 + 2t \right|}{2} + C$
$= \frac{1}{6} \log \left| t - 1 \right| + \frac{1}{2} \log \left| t + 1 \right| - \frac{2}{3}\log \left| 1 + 2t \right| + C$
$= \frac{1}{6} \log \left| \cos x - 1 \right| + \frac{1}{2} \log \left| \cos x + 1 \right| - \frac{2}{3} \log \left| 1 + 2 \cos x \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 61 | Page 17