Question

\[\int\frac{1}{3 + 4 \cot x} dx\]

Answer 1

\[\text{ Let I }= \int\frac{1}{3 + 4 \cot x}dx\]
\[ = \int\frac{1}{3 + \frac{4 \cos x}{\sin x}}dx\]
\[ = \int\frac{\sin x}{3 \sin x + 4 \cos x}dx\]
\[\text{ Let sin x = A }\left( 3 \sin x + 4 \cos x \right) + B \left( 3 \cos x - 4 \sin x \right) . . . (1)\]
\[ \Rightarrow \sin x = \left( 3A - 4B \right) \sin x + \left( 4A + 3B \right) \cos x\]
\[\text{By comparing the coefficients of both sides we get} , \]
\[3A - 4B = 1 . . . \left( 2 \right)\]
\[4A + 3B = 0 . . . \left( 3 \right)\]

Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get

\[9A - 12B + 16A + 12B = 3 + 0\]
\[ \Rightarrow 25A = 3\]
\[ \Rightarrow A = \frac{3}{25}\]
\[\text{  Putting value of A in eq }\left( 3 \right) \text{ we get,} \]
\[4 \times \frac{3}{25} + 3B = 0\]
\[ \Rightarrow 3B = - \frac{12}{25}\]
\[ \Rightarrow B = - \frac{4}{25}\]

\[\text{ Thus, by substituting the value of A and B in eq (1) we get }\]
\[I = \int\left[ \frac{\frac{3}{25}\left( 3 \sin x + 4 \cos x \right) - \frac{4}{25}\left( 3 \cos x - 4 \sin x \right)}{3 \sin x + 4 \cos x} \right]dx\]
\[ = \frac{3}{25}\int dx - \frac{4}{25}\int\left( \frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x} \right)dx\]
\[\text{  Putting   3  sin x + 4  cos x = t}\]
\[ \Rightarrow \left( 3 \cos x - 4 \sin x \right)dx = dt\]
\[ \therefore I = \frac{3}{25}\int dx - \frac{4}{25}\int\frac{dt}{t}\]
\[ = \frac{3}{25}x - \frac{4}{25} \text{ ln }\left| t \right| + C\]
\[ = \frac{3x}{25} - \frac{4}{25} \text{ ln }\left| 3 \sin x + 4 \cos x \right| + C\]