Question

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

Answer 1

\[\text{ Let I } = \int \frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)}dx\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]
\[ \Rightarrow I = \int \frac{\sec^2 x}{\left( \frac{\sin x - 2 \cos x}{\cos x} \right) \times \left( \frac{2 \sin x + \cos x}{\cos x} \right)}dx\]
\[ = \int \frac{\sec^2 x}{\left( \tan x - 2 \right) \left( 2 \tan x + 1 \right)}dx\]
\[\text{ Let tan x} = t\]
\[ \Rightarrow \sec^2 x \text{ dx } = dt\]
\[ \therefore I = \int \frac{dt}{\left( t - 2 \right) \left( 2t + 1 \right)}\]

\[ = \int \frac{dt}{2 t^2 + t - 4t - 2}\]
\[ = \int \frac{dt}{2 t^2 - 3t - 2}\]
\[ = \frac{1}{2}\int \frac{dt}{t^2 - \frac{3}{2}t - 1}\]
\[ = \frac{1}{2}\int \frac{dt}{t^2 - \frac{3}{2}t + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 - 1}\]
\[ = \frac{1}{2}\int \frac{dt}{\left( t - \frac{3}{4} \right)^2 - \frac{9}{16} - 1}\]
\[ = \frac{1}{2}\int \frac{dt}{\left( t - \frac{3}{4} \right)^2 - \left( \frac{5}{4} \right)^2}\]
\[ = \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \text{ log } \left| \frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}} \right| + C\]
\[ = \frac{1}{5} \text{ ln } \left| \frac{t - 2}{t + \frac{1}{2}} \right| + C\]
\[ = \frac{1}{5}\text{ ln } \left| \frac{\left( t - 2 \right)^2}{2t + 1} \right| + C\]
\[ = \frac{1}{5}\text{ln } \left| \frac{t - 2}{2t + 1} \right| + \frac{1}{5} \ln \left( 2 \right) + C\]
\[ = \frac{1}{5} \text{ ln } \left| \frac{t - 2}{2t + 1} \right| + \text{ C where C} = C + \frac{1}{5}\ln \left( 2 \right)\]
\[ = \frac{1}{5} \text{ ln } \left| \frac{\tan x - 2}{2 \tan x + 1} \right| + C\]