$\int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} dx$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{5}{13} x + \frac{12}{13} \ln | 3 \cos x + 2 \sin x | + C$

$- \frac{5}{13} x - \frac{12}{13} \ln | 3 \cos x + 2 \sin x | + C$

$- \frac{5}{13} x + \frac{12}{13} \ln | 3 \cos x + 2 \sin x | + C$

$- \frac{5}{13} x + \frac{12}{13} \ln | 2 \cos x + 3 \sin x | + C$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let

$\begin{array}{l} I = \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} dx \\ 3 \sin x + 2 \cos x = A \frac{d}{dx} (3 \cos x + 2 \sin x) + B (3 \cos x + 2 \sin x) \end{array}$

$\Rightarrow 3 \sin x + 2 \cos x = A (- 3 \sin x + 2 \cos x) + B (3 \cos x + 2 \sin x)$

On comparing the coefficients of sin x and cos x of both sides, we get

$\begin{array}{l} - 3 A + 2 B = 3 and 2 A + 3 B = 2 \\ \Rightarrow B = \frac{12}{13} and A = - \frac{5}{13} \\ ∴ I = \int \frac{B (- 3 \sin x + 2 \cos x) + A (3 \cos x + 2 \sin x)}{3 \cos x + 2 \sin x} dx \\ \Rightarrow I = A \int 1 dx + B \int \frac{- 3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} dx \\ = Ax + B \int \frac{dt}{t}, where t = 3 \cos x + 2 \sin x \\ = Ax + Bln | t | + C \\ = \frac{- 5}{13} x + \frac{12}{13} \ln | 3 \cos x + 2 \sin x | + C \end{array}$