Question

The value of $\int [1+\tan x\cdot \tan (x+\alpha \left)\right]dx$ is equal to

• A. $\cos \alpha \cdot \log |\frac{\sin x}{\sin (x+\alpha )}|+C$
• B. $\tan \alpha \cdot \log |\frac{\sin x}{\sin (x+\alpha )}|+C$
• C. $\cot \alpha \cdot \log |\frac{\sec (x+\alpha )}{\sec x}|+C$
• D. $\cot \alpha \cdot \log |\frac{\cos (x+\alpha )}{\cos x}|+C$

Answer 1

The correct option is C $\cot \alpha \cdot \log |\frac{\sec (x+\alpha )}{\sec x}|+C$

Let $I=\int [1+\tan x\tan (x+\alpha \left)\right]dx$

$[\because \tan (x+\alpha -x)=\frac{\tan (x+\alpha )-\tan x}{1+\tan x\tan (x+\alpha )}]$

so $I=\int \frac{\tan (x+\alpha )-\tan x}{\tan \alpha }dx$

$=\frac{1}{\tan \alpha }\left[\log \sec \right(x+\alpha )-\log \sec x]+c$

$=\cot \alpha \log |\frac{\sec (x+\alpha )}{\sec x}|+c$.