Q115 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
$\int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x=$
- Correct answerA$\frac{7}{15} \log |\cos 3 x+2 \sin 3 x|-\frac{4}{5} x+c$
- B$-\frac{4}{5} \log |\cos 3 x+2 \sin 3 x|+\frac{7 x}{5}+c$
- C$\frac{7}{5} \log |\cos 3 x+2 \sin 3 x|-\frac{4}{5} x+c$
- D$-\frac{8}{15} \log |\cos 3 x+2 \sin 3 x|+\frac{x}{5}+c$
Solution
$\frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x}$
$\begin{aligned}
& =\frac{A(\cos 3 x+2 \sin 3 x)+B(6 \cos 3 x-3 \sin 3 x)}{\cos 3 x+2 \sin 3 x} \\
& =\frac{(A+6 B) \cos 3 x+(2 A-3 B) \sin 3 x}{\cos 3 x+2 \sin 3 x} \\
& A+6 B=2 ; 2 A-3 B=-3
\end{aligned}$
After solving we get, $5 A=-4$ $\begin{aligned} & \Rightarrow A=\frac{-4}{5} \text { and } B=\frac{7}{15} \\ & =\frac{-4}{5} x+\frac{7}{15} \log |\cos 3 x+2 \sin 3 x|+c \end{aligned}$
After solving we get, $5 A=-4$ $\begin{aligned} & \Rightarrow A=\frac{-4}{5} \text { and } B=\frac{7}{15} \\ & =\frac{-4}{5} x+\frac{7}{15} \log |\cos 3 x+2 \sin 3 x|+c \end{aligned}$
Hence, the correct answer is option: 'A'.