Question

The value of $\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx$ is
(where $C$ is integration constant)

• A. $\frac{12}{13}x-\frac{5}{13}\ln |2\cos x-3\sin x|+C$
• B. $\frac{12}{13}x-\frac{5}{13}\ln |2\cos x+3\sin x|+C$
• C. $\frac{12}{13}x+\frac{5}{13}\ln |3\cos x-2\sin x|+C$
• D. $\frac{12}{13}x-\frac{5}{13}\ln |3\cos x+2\sin x|+C$

Answer 1

The correct option is D $\frac{12}{13}x-\frac{5}{13}\ln |3\cos x+2\sin x|+C$

$I=\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx$
Let,
$3\sin x+2\cos x=A(3\cos x+2\sin x)+B\frac{d}{dx}(3\cos x+2\sin x)\Rightarrow 3\sin x+2\cos x=A(3\cos x+2\sin x)+B(-3\sin x+2\cos x)\Rightarrow -3B+2A=3\text{and}2B+3A=2\Rightarrow A=\frac{12}{13}\text{and}B=-\frac{5}{13}\therefore I=\int \frac{A(3\cos x+2\sin x)+B(-3\sin x+2\cos x)}{3\cos x+2\sin x}dx=\frac{12}{13}\int 1\cdot dx-\frac{5}{13}\int \frac{-3\sin x+2\cos x}{3\cos x+2\sin x}dx=\frac{12}{13}x-\frac{5}{13}\ln |3\cos x+2\sin x|+C$