Question

\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]

Answer 1

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

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

  9A     -     12B   =   6 

16A   +    12B  =     12
_________________________

            25 A   = 18

 

` ⇒ A = 18/25 `
\[\text{ Putting  value of  A in  eq } \left( 2 \right) \text{   we get,} \]
\[ \Rightarrow B = \frac{1}{25}\]

\[\text{Thus, by substituting the values of A and B in eq (1) we get}, \]
\[I = \int\left\{ \frac{\frac{18}{25} \left( 3 \sin x + 4 \cos x \right) + \frac{1}{25}\left( 3 \cos x - 4 \sin x \right)}{3 \sin x + 4 \cos x} \right\}dx\]
\[ = \frac{18}{25}\int dx + \frac{1}{25}\int\left( \frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x} \right)dx\]
\[\text{ Putting 3 sin x + 4 cos x = t }\]
\[ \Rightarrow \left( 3 \cos x - 4 \sin x \right)dx = dt\]
\[ \therefore I = \frac{18}{25}x + \frac{1}{25}\int\frac{1}{t}dt\]
\[ = \frac{18x}{25} + \frac{1}{25} \text{ ln }\left| t \right| + C\]
\[ = \frac{18x}{25} + \frac{1}{25} \text{ ln }\left| 3 \sin x + 4 \cos\ x \right| + C\]