Question

If $\int \frac{3sinx+2cosx}{3cosx+2sinx}dx=ax+bln|2sinx+3cosx|+c$, then

• A. $a=-\frac{12}{13},b=\frac{15}{39}$
• B. $a=\frac{17}{13},b=\frac{6}{13}$
• C. $a=\frac{12}{13},b=\frac{15}{39}$
• D. $a=-\frac{17}{13},b=-\frac{1}{192}$

Answer 1

The correct option is C $a=\frac{12}{13},b=\frac{15}{39}$

Let $I=\int \frac{3sinx+2cosx}{2sinx+3cosx}dx$
Multiply numerator and denominator by ${\sec }^{2}x$
$\therefore I=\int \frac{2{\sec }^{2}x+3{\sec }^{2}x\tan x}{3{\sec }^{2}x+2{\sec }^{2}x\tan x}dx=\int \frac{(2+3\tan x){\sec }^{2}x}{(3+2\tan x)(1+{\tan }^{2}x)}dx$
Put $u=\tan x\Rightarrow du={\sec }^{2}xdx$
$I=\int \frac{3u+2}{(2u+3)(1+t^2)}du=\int (\frac{5u+12}{13(u^2+1)}-\frac{10}{13(2u+3)})du$
$=\frac{1}{13}\int (\frac{5u}{u^2+1}+\frac{12}{u^2+1})du-\frac{10}{13}\int \frac{1}{2u+3}du$
$=\frac{5}{26}\log (u^2+1)-\frac{5}{13}\log (2u+3)+\frac{12}{13}{\tan }^{-1}u+c$
$=\frac{1}{13}(12x-5\log (2sinx+3cosx))+c$