Question

Evaluate:$\int \frac{dx}{1+4\sin x+3\cos x}$

• A. $I=\frac{\sqrt{3}}{12\sqrt{2}}\log |\frac{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}-2\sqrt{2}}{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}+2\sqrt{2}}|+c$
• B. $I=\frac{-\sqrt{3}}{12\sqrt{2}}\log |\frac{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}-2\sqrt{2}}{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}+2\sqrt{2}}|+c$
• C. $I=\frac{-\sqrt{3}}{12\sqrt{2}}\log |\frac{\sqrt{3}\tan \frac{x}{2}+\sqrt{3}-2\sqrt{2}}{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}+2\sqrt{2}}|+c$
• D. None of these

Answer 1

Answer: (B)

$I=\frac{-\sqrt{3}}{12\sqrt{2}}\log |\frac{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}-2\sqrt{2}}{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}+2\sqrt{2}}|+c$

Putting $\sin x=\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}$ and $\cos x=\frac{2\tan \left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}$ , we have

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

$=\int \frac{1}{1+4\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}+3\frac{2\tan \left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}}dx$

$=\int \frac{1+{\tan }^2\left(\frac{x}{2}\right)}{5+6\tan \left(\frac{x}{2}\right)-3{\tan }^2\left(\frac{x}{2}\right)}dx$

$=\int \frac{{\sec }^2\left(\frac{x}{2}\right)dx}{5+6\tan \left(\frac{x}{2}\right)-3{\tan }^2\left(\frac{x}{2}\right)}$

Let $\tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}dx=dt$

Thus above integral can be written as

$\int \frac{dt}{5+6t-3t^2}$

$=\frac{-1}{3}\int \frac{dt}{t^2-2t-\frac{5}{3}}$

$=\frac{-1}{3}\int \frac{dt}{(t-1{)}^2-{\left(2\sqrt{\frac{2}{3}}\right)}^2}$

$=\frac{-\sqrt{3}}{12\sqrt{2}}\log \left|\frac{\sqrt{3}t-\sqrt{3}-2\sqrt{2}}{\sqrt{3}t-\sqrt{3}+2\sqrt{2}}\right|+c$ ............ Using $\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+c$

Putting $t=\tan \frac{x}{2}$ ,we have $I=\frac{-\sqrt{3}}{12\sqrt{2}}\log \left|\frac{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}-2\sqrt{2}}{\sqrt{3}\tan \frac{x}{2}-\sqrt{3}+2\sqrt{2}}\right|+c$