Question

$\int \frac{2\tan x+3}{3\tan x+4}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{2\tan x+3}{3\tan x+4}\right)dx \\ =\int \left(\frac{{\displaystyle \frac{2\sin x}{\cos x}}+3}{{\displaystyle \frac{3\sin x}{\cos x}}+4}\right)dx \\ =\int \left(\frac{2\sin x+3\cos x}{3\sin x+4\cos x}\right)dx \\ \mathrm{Let}\left(2\sin x+3\cos x\right)=A\left(3\sin x+4\cos x\right)+B\left(3\cos x-4\sin x\right)....\left(1\right) \\ \Rightarrow 2\sin x+3\cos x=\left(3A-4B\right)\sin x+\left(4A+3B\right)\cos x \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 3A-4B=2...\left(2\right) \\ 4A+3B=3...\left(3\right) \end{gathered}$$

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

$$\begin{gathered} \underline{\begin{array}{ccccc}9A& -& 12B& =& 6\\ 16A& +& 12B& =& 12\end{array}} \\ \begin{array}{ccccc}& & 25A& =& 18\end{array} \\ \Rightarrow A=\frac{18}{25} \\ \mathrm{Putting}\mathrm{value}\mathrm{of}A\mathrm{in}\mathrm{eq}\left(2\right)\mathrm{we}\mathrm{get}, \\ \Rightarrow B=\frac{1}{25} \end{gathered}$$
$$\begin{gathered} \mathrm{Thus},\mathrm{by}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}A\mathrm{and}B\mathrm{in}\mathrm{eq}\left(1\right)\mathrm{we}\mathrm{get}, \\ I=\int \left\{\frac{{\displaystyle \frac{18}{25}}\left(3\sin x+4\cos x\right)+{\displaystyle \frac{1}{25}}\left(3\cos x-4\sin x\right)}{3\sin x+4\cos x}\right\}dx \\ =\frac{18}{25}\int dx+\frac{1}{25}\int \left(\frac{3\cos x-4\sin x}{3\sin x+4\cos x}\right)dx \\ \mathrm{Putting}3\sin x+4\cos x=t \\ \Rightarrow \left(3\cos x-4\sin x\right)dx=dt \\ \therefore I=\frac{18}{25}x+\frac{1}{25}\int \frac{1}{t}dt \\ =\frac{18x}{25}+\frac{1}{25}\ln \left|t\right|+\mathrm{C} \\ =\frac{18x}{25}+\frac{1}{25}\ln \left|3\sin x+4\cos x\right|+\mathrm{C} \end{gathered}$$