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

Sum
$\int\frac{\sin 2x}{\left( 1 + \sin x \right) \left( 2 + \sin x \right)} dx$

#### Solution

We have,
$I = \int\frac{\sin 2x dx}{\left( 1 + \sin x \right) \left( 2 + \sin x \right)}$
$= \int\frac{2 \sin x \cos x dx}{\left( 1 + \sin x \right) \left( 2 + \sin x \right)}$
Putting sin x = t

$\Rightarrow \cos x dx = dt$
$\therefore I = \int\frac{2t dt}{\left( 1 + t \right) \left( 2 + t \right)}$
$= 2\int\frac{t dt}{\left( 1 + t \right) \left( 2 + t \right)}$
$\text{Let }\frac{t}{\left( 1 + t \right) \left( 2 + t \right)} = \frac{A}{1 + t} + \frac{B}{2 + t}$
$\Rightarrow \frac{t}{\left( 1 + t \right) \left( 2 + t \right)} = \frac{A \left( 2 + t \right) + B \left( 1 + t \right)}{\left( 1 + t \right) \left( 2 + t \right)}$
$\Rightarrow t = A \left( 2 + t \right) + B \left( 1 + t \right)$
Putting 2 + t = 0

$\Rightarrow t = - 2$
$- 2 = A \times 0 + B \left( - 2 + 1 \right)$
$\Rightarrow - 2 = B \left( - 1 \right)$
$\Rightarrow B = 2$
$\text{Let }t + 1 = 0$
$t = - 1$
$\Rightarrow - 1 = A \left( - 1 + 2 \right) + B \times 0$
$A = - 1$
$\therefore I = 2\int\left( \frac{- 1}{t + 1} + \frac{2}{t + 2} \right)dt$
$= 2 \left[ - \log \left| t + 1 \right| + 2 \log \left| t + 2 \right| \right] + C$
$= 4 \log \left| t + 2 \right| - 2 \log \left| t + 1 \right| + C$
$= \log \left| \frac{\left( t + 2 \right)^4}{\left( t + 1 \right)^2} \right| + C$
$= \log \left| \frac{\left( \sin x + 2 \right)^4}{\left( \sin x + 1 \right)^2} \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 11 | Page 176