Question

$\int \frac{3+4\sin x+2\cos x}{3+2\sin x+\cos x}dx.$

• A. $2x+3{\tan }^{-1}(\tan \frac{x}{2}+1)+c$
• B. $2x-3{\tan }^{-1}(\tan \frac{x}{2}+1)+c$
• C. $2x-6{\tan }^{-1}(\tan \frac{x}{2}+1)+c$
• D. $x-3{\tan }^{-1}(\tan \frac{x}{2}+1)+c$

Answer 1

The correct option is B $2x-3{\tan }^{-1}(\tan \frac{x}{2}+1)+c$

Let $I=\int \frac{3+4\sin x+2\cos x}{3+2\sin x+\cos x}dx$
$=\int \frac{6+4\sin x+2\cos x-3}{3+2\sin x+\cos x}dx$
$=\int 2dx-3\int \frac{1}{3+2\sin x+\cos x}dx$
$=2x-3\int \frac{1+{\tan }^2\frac{x}{2}}{3(1+{\tan }^2\frac{x}{2})+4\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}}dx$
$=2x-3\int \frac{{\sec }^2\frac{x}{2}}{2{\tan }^2\frac{x}{2}+4\tan \frac{x}{2}+4}dx$
Substitute $\tan \frac{x}{2}=t$
$\Rightarrow {\sec }^2\frac{x}{2}\frac{1}{2}dx=dt$
$=2x-3\int \frac{2}{2t^2+4t+4}dt$
$I=2x-3\int \frac{1}{{(t+1)}^2+1}dt$
$I=2x-3{\tan }^{-1}(t+1)+C$
$\Rightarrow I=2x-3{\tan }^{-1}(\tan \frac{x}{2}+1)+c$