Question

$\int_0^{\pi / 4} \log (1+\tan x) \text{d}x=$

• A. $\frac{\pi}{16} \log 2$
• B. $\frac{\pi}{4} \log 2$
• C. $\frac{\pi}{8} \log 2$
• D. $\pi \log 2$

Hint:

This is one of the most famous standard results in calculus. Memorizing $\int_0^{\pi/4} \log(1+\tan x)\text{d}x = \frac{\pi}{8}\log 2$ will save you several minutes on competitive exams!

Answer 1

The Correct Option is C

Step 1: Apply the king property.
Let $I = \int_0^{\pi/4}\log(1+\tan x)\,dx$. Replace $x$ by $\dfrac{\pi}{4} - x$.

Step 2: Simplify the new integrand.
$1 + \tan\left(\dfrac{\pi}{4} - x\right) = \dfrac{2}{1 + \tan x}$, so $I = \int_0^{\pi/4}\log 2\,dx - I$.

Step 3: Solve for I.
$2I = \dfrac{\pi}{4}\log 2$, so $I = \dfrac{\pi}{8}\log 2$.
\[ \boxed{\dfrac{\pi}{8}\log 2,\ \text{option 3}} \]