Question

\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{  dx }\]

Answer 1

\[\text{ Let I} = \int\left( \frac{8 \cot x + 1}{3 \cot x + 2} \right)dx\]
\[ = \int\left( \frac{8 \frac{\cos x}{\sin x} + 1}{\frac{3 \cos x}{\sin x} + 2} \right)dx\]
\[ = \int\left( \frac{8 \cos x + \sin x}{3 \cos x + 2 \sin x} \right)dx\]
\[\text{ Now, let 8  cos x + sin x = A }\left( 3 \cos x + 2 \sin x \right) + B \left( - 3 \sin x + 2 \cos x \right) . . . (1) \]
\[ \Rightarrow 8 \cos x + \sin x = 3A \cos x + 2A \sin x - 3B \sin x + 2B \cos x \]
\[ \Rightarrow 8 \cos x + \sin x = \left( 3A + 2B \right) \cos x + \left( 2A - 3B \right) \sin x \]
\[\text{Equating the coefficients of like terms we get}, \]
\[2A - 3B = 1 . . . \left( 2 \right)\]
\[3A + 2B = 8 . . . \left( 3 \right)\]

Solving eq (2) and  eq (3) we get,
A = 2, B = 1
Thus, by substituting the values of A and B in eq (1) we get ,

\[I = \int\left[ \frac{2 \left( 3 \cos x + 2 \sin x \right) + 1\left( - 3 \sin x + 2 \cos x \right)}{\left( 3 \cos x + 2 \sin x \right)} \right]dx\]
\[ = 2\int\left( \frac{3 \cos x + 2 \sin x}{3 \cos x + 2 \sin x} \right)dx + \int\left( \frac{- 3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \right)dx\]
\[ = 2\int dx + \int\left( \frac{- 3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \right)dx\]
\[\text{ Putting   3 cos x + 2 sin x = t }\]
\[ \Rightarrow \left( \text{  - 3  sin x + 2 cos x} \right)dx = dt \]
\[ \therefore I = 2\int dx + \int\frac{1}{t}dt\]
\[ = 2x + \text{ ln }\left| t \right| + C\]
\[ = 2x + \text{ ln }\left| 3 \cos x + 2 \sin x \right| + C\]