Question

What is $\int {\tan }^{-1}(\sec x+\tan x)dx$ equal to?

• A. $\frac{\pi x}{4}+\frac{x^2}{4}+c$
• B. $\frac{\pi x}{2}+\frac{x^2}{4}+c$
• C. $\frac{\pi x}{4}+\frac{\pi x^2}{4}+c$
• D. $\frac{\pi x}{4}-\frac{x^2}{4}+c$

Answer 1

Answer: (A)

$\frac{\pi x}{4}+\frac{x^2}{4}+c$

$\sec x+\tan x=\frac{1+\sin x}{\cos x}=\frac{(\cos \frac{x}{2}+\sin \frac{x}{2}{)}^2}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}=\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}=\tan (\frac{\pi }{4}+\frac{x}{2})$

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

So correct answer will be option A.