Question

\[\int\frac{\sin 2x}{\left( 1 + \sin x \right) \left( 2 + \sin x \right)} dx\]

Answer 1

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

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