Question

\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]

Answer 1

We have,
\[ I = \int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\text{Let, }\sin x = t\]
\[ \Rightarrow \cos x dx = dt\]
\[\text{Now, integration becomes}, \]
\[I = \int\frac{dt}{\left( 1 - t \right)^3 \left( 2 + t \right)} \]
\[ = - \int\frac{dt}{\left( t - 1 \right)^3 \left( t + 2 \right)} \]
\[\text{Let, }\frac{1}{\left( t - 1 \right)^3 \left( t + 2 \right)} = \frac{A}{\left( t - 1 \right)} + \frac{B}{\left( t - 1 \right)^2} + \frac{C}{\left( t - 1 \right)^3} + \frac{D}{\left( t + 2 \right)} ................(1)\]
\[ \Rightarrow 1 = A \left( t - 1 \right)^2 \left( t + 2 \right) + B\left( t - 1 \right)\left( t + 2 \right) + C\left( t + 2 \right) + D \left( t - 1 \right)^3 ....................(2)\]

\[\text{Putting t = 1 in (2), we get}\]
\[1 = 3C\]
\[ \Rightarrow C = \frac{1}{3}\]
\[\text{Putting t = - 2 in (2), we get}\]
\[1 = D \left( - 2 - 1 \right)^3 \]
\[ \Rightarrow 1 = - 27D\]
\[ \Rightarrow D = \frac{- 1}{27}\]
\[\text{Putting t = 0 in (2), we get}\]
\[1 = 2A - 2B + 2C - D\]
\[ \Rightarrow 1 = 2A - 2B + \frac{2}{3} + \frac{1}{27}\]
\[ \Rightarrow 2A - 2B = \frac{8}{27}\]
\[ \Rightarrow A - B = \frac{4}{27}\]
\[\text{Putting t = 2 in (2), we get}\]
\[1 = 4A + 4B + 4C + D\]
\[ \Rightarrow 1 = 4A + 4B + \frac{4}{3} - \frac{1}{27}\]
\[ \Rightarrow A + B = - \frac{2}{27}\]
\[Now, A - B = \frac{4}{27}\text{ and }A + B = - \frac{2}{27} \Rightarrow A = \frac{1}{27}\text{ and }B = \frac{- 1}{9}\]

\[\text{Substituting the values of A, B, C and D in (1), we get}\]
\[\frac{1}{\left( t - 1 \right)^3 \left( t + 2 \right)} = \frac{1}{27\left( t - 1 \right)} - \frac{1}{9 \left( t - 1 \right)^2} + \frac{1}{3 \left( t - 1 \right)^3} + \frac{- 1}{27\left( t + 2 \right)}\]
\[\text{Now, integration becomes}\]
\[ I = - \int\left[ \frac{1}{27\left( t - 1 \right)} - \frac{1}{9 \left( t - 1 \right)^2} + \frac{1}{3 \left( t - 1 \right)^3} + \frac{- 1}{27\left( t + 2 \right)} \right]dt\]
\[ = - \left[ \frac{1}{27}\log \left| t - 1 \right| + \frac{1}{9\left( t - 1 \right)} - \frac{1}{6 \left( t - 1 \right)^2} - \frac{1}{27}\log \left| t + 2 \right| \right] + C\]
\[ = - \frac{1}{27}\log \left| \sin x - 1 \right| - \frac{1}{9\left( \sin x - 1 \right)} + \frac{1}{6 \left( \sin x - 1 \right)^2} + \frac{1}{27}\log \left| \sin x + 2 \right| + C\]
\[ = - \frac{1}{27}\log \left| 1 - \sin x \right| + \frac{1}{9\left( 1 - \sin x \right)} + \frac{1}{6 \left( 1 - \sin x \right)^2} + \frac{1}{27}\log \left| 2 + \sin x \right| + C\]