Question

\[\int\frac{\sin x + 2 \cos x}{2 \sin x + \cos x} \text{ dx }\]

Answer 1

\[\int\left( \frac{\sin x + 2 \cos x}{2 \sin x + \cos x} \right)dx\]
\[\text{ Let  sin x + 2  cos x = A } \frac{d}{dx} \left( \text{ 2 sin x + cos x} \right) + \text{ B }\left( \text{ 2  sin x + cos x} \right)\]
\[ \Rightarrow \sin x + 2 \cos x = A \left( 2 \cos x - \sin x \right) + \text{ 2  B sin x + B  cos  x}\]
\[ \Rightarrow \sin x + 2 \cos x = \left( \text{ 2  A + B }\right) \cos x + \left( 2 B - A \right) \sin x\]
\[\text{Equating coefficients of like terms}\]
\[ \Rightarrow \text{ 2  A + B = 2} . . . \left( 1 \right)\]
\[ \Rightarrow - A + 2B = 1 . . . \left( 2 \right)\]
\[\text{Multiplying eq} \left( 2 \right) \text{by 2 and adding it to eq} \left( 1 \right) \text{we get}, \]
\[\text{ 5  B = 4 }\]
\[ \Rightarrow B = \frac{4}{5}\]
\[\text{  Putting  B }= \frac{4}{5} \text{ in eq }\left( 1 \right) \text{ we get,} \]
\[2 A + \frac{4}{5} = 2\]
\[ \Rightarrow A = \frac{3}{5}\]
\[ \therefore \int\left( \frac{\sin x + 2 \cos x}{2 \sin x + \cos x} \right)dx = \int\left[ \frac{\frac{3}{5} \left( 2 \cos x - \sin x \right)}{2 \sin x + \cos x} \right]dx + \frac{4}{5}\int\frac{\left( 2 \sin x + \cos x \right)}{\left( 2 \sin x + \cos x \right)}dx\]
\[ = \frac{3}{5}\int\left( \frac{2 \cos x - \sin x}{2 \sin x + \cos x} \right)dx + \frac{4}{5}\int dx\]
\[\text{ Putting 2  sin x + cos x = t }\]
\[ \Rightarrow \left( 2 \cos x - \sin x \right) dx = dt\]
\[ \therefore I = \frac{3}{5}\int\frac{dt}{t} + \frac{4}{5}\int dx\]
\[ = \frac{3}{5} \text{ ln  }\left| t \right| + \frac{4x}{5} + C\]
\[ = \frac{3}{5} \text{ ln } \left| 2 \sin x + \cos x \right| + \frac{4x}{5} + C ...............\left[ \because t = 2 \sin x + \cos x \right]\]