Question

$\int \frac{dx}{\cot \frac{x}{2}\cot \frac{x}{3}\cot \frac{x}{6}}$ =

• A. $\ln \left|\sec \frac{x}{2}\sec \frac{x}{3}\sec \frac{x}{6}\right|+c$
• B. $\ln \left|{\sec }^2\frac{x}{2}{\sec }^3\frac{x}{3}{\sec }^6\frac{x}{6}\right|+c$
• C. $\ln \left|{\sec }^2\frac{x}{2}{\cos }^3\frac{x}{3}{\cos }^6\frac{x}{6}\right|+c$
• D. $\ln \left|{\tan }^2\frac{x}{2}{\tan }^3\frac{x}{3}{\sec }^5\frac{x}{6}\right|+c$

Answer 1

The correct option is C $\ln \left|{\sec }^2\frac{x}{2}{\cos }^3\frac{x}{3}{\cos }^6\frac{x}{6}\right|+c$

$\int \frac{dx}{\cot \frac{x}{2}\cot \frac{x}{3}\cot \frac{x}{6}}$

$I=\int \tan \frac{x}{2}\tan \frac{x}{3}\tan \frac{x}{6}dn$

$\Rightarrow \tan \frac{x}{6}=\tan (\frac{x}{2}-\frac{x}{3})$ $\left[\tan \right(AB)=\frac{\tan A-\tan B}{1+\tan A+\tan B}]$

$tan\frac{x}{6}=\frac{\tan x/2+\tan x/3}{1+\tan x/2\tan x/3}$

$\Rightarrow \tan \frac{x}{6}(1+\tan \frac{x}{2}\tan \frac{x}{3})=\tan \frac{x}{2}-\tan \frac{x}{3}$

$\Rightarrow \tan \frac{x}{6}+\tan \frac{x}{2}\tan \frac{x}{3}\tan \frac{x}{6}=\tan \frac{x}{2}-\tan \frac{x}{3}$

Integrating both sides we get

$\tan \frac{x}{6}dx+\int \tan \frac{x}{2}\tan \frac{x}{3}\tan \frac{x}{6}dn=\int \tan \frac{x}{2}dx-\int \tan \frac{x}{3}dn$

$I=\int \tan \frac{x}{2}dx-\int \tan \frac{x}{3}dn-\int \tan \frac{x}{6}dn$ [from equation]$=2log\left|\sec \frac{x}{2}\right|-3log\left|\sec \frac{x}{3}\right|-6log\left|\sec \frac{x}{6}\right|+C$

$=log\left|{\sec }^2\frac{x}{2}\right|+log\left|{\cos }^3\frac{x}{3}\right|+log\left|{\cos }^6\frac{x}{6}\right|+C$

$I=log|{\sec }^2\frac{x}{2}+{\cos }^3\frac{x}{3}+{\cos }^6\frac{x}{6}|+C$