If $\int \frac{3 \sin x+2 \cos x}{3 \cos x+2 \sin x} d x=a x+b \log |3 \cos x+2 \sin x|+C$, then
Answer & explanation
Correct answer: option 2
Let
$I=\int \frac{3 \sin x+2 \cos x}{3 \cos x+2 \sin x} d x$
Let $3 \sin x+2 \cos x$
$=\lambda \frac{d}{d x}(3 \cos x+2 \sin x)+\mu(3 \cos x+2 \sin x)$
$\Rightarrow 3 \sin x+2 \cos x =\lambda(-3 \sin x+2 \cos x)+\mu(3 \cos x+2 \sin x)$
Comparing the coefficients of $\sin x$ and $\cos x$ on both sides, we get
$-3 \lambda+2 \mu=3$ and $2 \lambda+3 \mu=2$
$\Rightarrow \lambda=-5 / 13$ and $\mu=12 / 13$
∴ $I=\int \frac{-\left(\frac{5}{13}\right)(-3 \sin x+2 \cos x)+\left(\frac{12}{13}\right)(3 \cos x+2 \sin x)}{3 \cos x+2 \sin x} d x$
$\Rightarrow I=\frac{12}{13} \int 1 . d x-\frac{5}{13} \int \frac{-3 \sin x+2 \cos x}{3 \cos x+2 \sin x} d x $
$\Rightarrow I=\frac{12}{13} \int 1 . d x-\frac{5}{13} \int \frac{1}{3 \cos x+2 \sin x} d(3 \cos x+2 \sin x)$
$\Rightarrow I=\frac{12}{13} x-\frac{5}{13} \log |3 \cos x+2 \sin x|+C$
Hence, $a=\frac{12}{13}$ and $b=\frac{-5}{13}$