Question

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

Answer 1

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