Question

Evaluate $\int \frac{dx}{1+4sinx+3cosx}$

• A. $\frac{1}{2\sqrt{6}}\left|\frac{\sqrt{6}+tanx/2-2}{\sqrt{6}-tanx/2+2}\right|+c$
• B. $\frac{1}{3\sqrt{6}}\left|\frac{\sqrt{6}+tanx/2-2}{\sqrt{6}+tanx/2+2}\right|+c$
• C. $\frac{1}{3\sqrt{6}}\left|\frac{\sqrt{6}+tanx/2+2}{\sqrt{6}+tanx/2+2}\right|+c$
• D. $\frac{1}{2\sqrt{6}}\left|\frac{\sqrt{6}+tanx/2+2}{\sqrt{6}-tanx/2+2}\right|+c$

Answer 1

The correct option is A $\frac{1}{2\sqrt{6}}\left|\frac{\sqrt{6}+tanx/2-2}{\sqrt{6}-tanx/2+2}\right|+c$

Let $I=\int \frac{1}{1+4\sin x+3\cos x}dx$

Substitute $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
$I=\int \frac{dt}{(t^2+1)(\frac{8t}{t^2+1}+3\frac{1-t^2}{t^2+1}+1)}$
$=\int \frac{1}{-t^2+4t+2}dt=\int \frac{1}{6-{(t-2)}^2}dt$
$=\frac{1}{2\sqrt{6}}\log \left(\frac{\sqrt{6}+t-2}{\sqrt{6}-t+2}\right)=\frac{1}{2\sqrt{6}}\log \left(\frac{\sqrt{6}+\tan \frac{x}{2}-2}{\sqrt{6}-\tan \frac{x}{2}+2}\right)$