Question

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

Answer 1

\[\text{  Let I }= \int\frac{dx}{4 + 3 \tan x}\]
\[ = \int\frac{dx}{4 + \frac{3 \sin x}{\cos x}}\]
\[ = \int\frac{\text{ cos x } dx}{4 \cos x + 3 \sin x}\]
\[\text{ Consider,} \]
\[\cos x = A \left( 4 \cos x + 3 \sin x \right) + B\frac{d}{dx}\left( 4 \cos x + 3 \sin x \right)\]
\[ \Rightarrow \cos x = A \left( 4 \cos x + 3 \sin x \right) + B \left( - 4 \sin x + 3 \cos x \right)\]
\[ \Rightarrow \cos x = \left( 4A + 3B \right) \cos x + \left( 3A - 4B \right) \sin x\]
\[\text{ Equating the coefficients of like terms }\]
\[4A + 3B = 1 . . . . . \left( 1 \right)\]
\[3A - 4B = 0 . . . . . \left( 2 \right)\]

Solving (1) and (2), we get

\[A = \frac{4}{25} \text{ and B }= \frac{3}{25}\]

\[\int\left[ \frac{\frac{4}{25}\left( 4 \cos x + 3 \sin x \right) + \left( - 4 \sin x + 3 \cos x \right)\frac{3}{25}}{4 \cos x + 3 \sin x} \right]dx\]
\[ = \frac{4}{25}\int dx + \frac{3}{25}\int\left( \frac{- 4 \sin x + 3 \cos x}{4 \cos x + 3 \sin x} \right)dx\]
\[\text{ let 4  cos x + 3 sin x = t}\]
\[ \Rightarrow \left( - 4 \sin x + 3 \cos x \right)dx = dt\]
\[\text{ Then, }\]
\[I = \frac{4}{25}\int dx + \frac{3}{25}\int\frac{dt}{t}\]
\[ = \frac{4x}{25} + \frac{3}{25} \text{  log }\left| t \right| + C\]
\[ = \frac{4x}{25} + \frac{3}{25} \text{ log }\left| 4 \cos x + 3 \sin x \right| + C\]