Question

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

• A. $\frac{6}{5}x+\frac{3}{5}\ln |\sin x+2\cos x+3|-\frac{8}{5}{\tan }^{-1}\left(\frac{1}{2}(\tan \frac{x}{2}+1)\right)+C$
• B. $\frac{3}{5}x+\frac{6}{5}\ln |\sin x+2\cos x+3|-\frac{8}{5}{\tan }^{-1}\left(\frac{1}{2}(\tan \frac{x}{2}+1)\right)+C$
• C. $\frac{6}{5}x-\frac{3}{5}\ln |2\sin x+\cos x+3|+\frac{8}{5}{\tan }^{-1}\left(\frac{1}{2}(\tan \frac{x}{2}+1)\right)+C$
• D. $\frac{3}{5}x+\frac{6}{5}\ln |2\sin x+\cos x+3|-\frac{8}{5}{\tan }^{-1}\left(\frac{1}{2}(\tan \frac{x}{2}+1)\right)+C$

Answer 1

The correct option is A $\frac{6}{5}x+\frac{3}{5}\ln |\sin x+2\cos x+3|-\frac{8}{5}{\tan }^{-1}\left(\frac{1}{2}(\tan \frac{x}{2}+1)\right)+C$

$I=\int \frac{3\cos x+2}{\sin x+2\cos x+3}dx$
Let,
$3\cos x+2=A(\sin x+2\cos x+3)+B(\cos x-2\sin x)+D$
Comparing the coefficients of $\sin x,\cos x$ and constant term on both sides, we get
$A-2B=0,2A+B=3,3A+D=2\Rightarrow A=\frac{6}{5},B=\frac{3}{5}\text{and}D=-\frac{8}{5}\therefore I=\int \frac{A(\sin x+2\cos x+3)+B(\cos x-2\sin x)+D}{\sin x+2\cos x+3}dx\Rightarrow I=A\int dx+B\int \frac{\cos x-2\sin x}{\sin x+2\cos x+3}dx+D\int \frac{1}{\sin x+2\cos x+3}dx\Rightarrow I=Ax+B\ln |\sin x+2\cos x+3|+D\cdot I_1\text{where}{I}_1=\int \frac{1}{\sin x+2\cos x+3}dx{I}_1=\int \frac{1}{\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\frac{2(1-{\tan }^2\frac{x}{2})}{1+{\tan }^2\frac{x}{2}}+3}dx[\because \sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}},\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}]=\int \frac{1+{\tan }^2\frac{x}{2}}{2\tan \frac{x}{2}+2-2{\tan }^2\frac{x}{2}+3(1+{\tan }^2\frac{x}{2})}dx=\int \frac{{\sec }^2\frac{x}{2}}{{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}+5}dx$

Putting $\tan \frac{x}{2}=t$
$\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}=dt\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$
Now,
$I_1=\int \frac{2dt}{t^2+2t+5}=2\int \frac{dt}{(t+1{)}^2+2^2}=\frac{2}{2}{\tan }^{-1}\left(\frac{t+1}{2}\right)+C={\tan }^{-1}\left[\frac{1}{2}(\tan \frac{x}{2}+1)\right]+C\therefore I=\frac{6}{5}x+\frac{3}{5}\ln |\sin x+2\cos x+3|-\frac{8}{5}{\tan }^{-1}\left(\frac{1}{2}(\tan \frac{x}{2}+1)\right)+C$