Question

Solve $\int \frac{co\sec \left({\tan }^{-1}x\right)}{1+x^2}dx$

• A. $\log \tan \left(\frac{1}{2}\tan x\right)+c$
• B. $\log \tan \left(\frac{1}{2}{\tan }^{-1}x\right)+c$
• C. $\tan \left(\frac{1}{2}{\tan }^{-1}x\right)+c$
• D. $\log \left(\frac{1}{2}{\tan }^{-1}x\right)+c$

Answer 1

The correct option is B $\log \tan \left(\frac{1}{2}{\tan }^{-1}x\right)+c$

$I=\int \frac{co\sec \left({\tan }^{-1}x\right)}{1+x^2}dx.$ Put ${\tan }^{-1}x=t.\therefore \frac{1}{1+x^2}dx=dt$ $\therefore I=\int co\sec tdt=\log \tan \left(\frac{t}{2}\right)+c=\log \tan \left(\frac{1}{2}{\tan }^{-1}x\right)+c$