Question

Evaluate

$\int \frac{3sinx+2cosx}{3cosx+2sinx}dx$

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

Answer 1

The correct option is B $\frac{12}{13}x-\frac{5}{13}ln|3cosx+2sinx|dx$

Let $I=\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx$
Multiply numerator and denominator by ${\sec }^{3}x$
$I=\int \frac{2{\sec }^{2}x+3\tan x{\sec }^{2}x}{3{\sec }^{2}x+2\tan x{\sec }^{2}x}dx=\int \frac{(3\tan x+2){\sec }^{2}x}{(2\tan x+3)({\tan }^{2}x+1)}dx$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$I=\int \frac{3t+2}{(2t+3)(t^2+1)}dt=\int (\frac{5t+12}{13(t^2+1)}-\frac{10}{13(2t+3)})dt$
$=\frac{1}{13}\int (\frac{5t}{t^2+1}+\frac{12}{t^2+1})dt-\frac{10}{13}\int \frac{1}{2t+3}dt$
$=\frac{12}{13}{\tan }^{-1}t+\frac{5}{13}\log (t^2+1)-\frac{5}{13}\log (2t+3)$
$=\frac{12}{13}x-\frac{5}{13}\log (2\sin x+3\cos x)$