Question

$\int \tan^{-1}(\sec x + \tan x) dx =$

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

Hint:

The identity $\sec x + \tan x = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$ is highly useful across calculus. Memorizing this transformation allows you to skip straight to integrating $\frac{\pi}{4} + \frac{x}{2}$ in seconds!

Answer 1

The Correct Option is A

Step 1: Simplify the inside term.
Write $\sec x + \tan x = \dfrac{1 + \sin x}{\cos x}$. Using half angle forms this becomes $\dfrac{1 + \tan(x/2)}{1 - \tan(x/2)}$.

Step 2: Recognise the tangent.
That fraction equals $\tan\left(\dfrac{\pi}{4} + \dfrac{x}{2}\right)$, so $\tan^{-1}$ of it is just $\dfrac{\pi}{4} + \dfrac{x}{2}$.

Step 3: Integrate.
\[ \int\left(\frac{\pi}{4} + \frac{x}{2}\right)dx = \frac{\pi x}{4} + \frac{x^2}{4} + c \]
\[ \boxed{\dfrac{\pi x}{4} + \dfrac{x^2}{4} + c,\ \text{option 1}} \]