Question

If $I=\int [1+\cot (x-\alpha \left)\cot \right(x+\alpha \left)\right]dx$, then I equsl

• A. $\log \left|\frac{\cot x-\cot \alpha }{\cot x+\cot \alpha }\right|+C$
• B. $\cot 2\alpha \log \left|\frac{1-\cot x\tan \alpha }{1+\cot x\tan \alpha }\right|+C$
• C. $cosec2\alpha \log \left|\frac{\tan x-\cot \alpha }{\tan x+\cot \alpha }\right|+C$
• D. $\log |\tan x|-x\log |\tan \alpha |+C$

Answer 1

Answer: (B)

$\cot 2\alpha \log \left|\frac{1-\cot x\tan \alpha }{1+\cot x\tan \alpha }\right|+C$

$I=\int [1+\cot (x-\alpha )\cot (x+\alpha )]dx=\int [1+\frac{1}{\tan (x-\alpha )\tan (x+\alpha )}]dx=\int [1+\frac{(1+\tan x\tan \alpha )(1-\tan x\tan \alpha )}{(\tan x-\tan \alpha )(\tan x+\tan \alpha )}]dx=\int \left[\frac{{\tan }^{2}x-{\tan }^2\alpha +1-{\tan }^{2}x{\tan }^2\alpha }{{\tan }^{2}x-{\tan }^2\alpha }\right]dx=\int \frac{(1-{\tan }^2\alpha ){\sec }^{2}x}{{\tan }^{2}x-{\tan }^2\alpha }dx=\frac{1-{\tan }^2\alpha }{2\tan \alpha }\log \left|\frac{\tan x-\tan \alpha }{\tan x+\tan \alpha }\right|+c=\csc 2\alpha \log \left|\frac{\tan x-\tan \alpha }{\tan x+\tan \alpha }\right|+c$