Definite Integration
Substitution — Trigonometric
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^{\pi/4}\ln(1+\tan x)\,dx\) [JEE Advanced 2001]</p>
\pi/8 \cdot ln2
\pi/4 \cdot ln2
\pi ln2
\pi/2 \cdot ln2
Step-by-Step Solution
Key Concept: King: I = \int_0^(\pi/4) ln(1+tan(\pi/4-x))dx = \int_0^(\pi/4) ln(2/(1+tanx))dx = \int(ln2 - ln(1+tanx))dx. Add: 2I = (\pi/4)ln2 \to I = (\pi/8)ln2.
<div class='solution'>
<p>King ($x\to\pi/4-x$): $\tan(\pi/4-x)=\frac{1-\tan x}{1+\tan x}$.</p>
<p>$1+\tan(\pi/4-x)=1+\frac{1-\tan x}{1+\tan x}=\frac{2}{1+\tan x}$.</p>
<p>$$I=\int_0^{\pi/4}\ln\frac{2}{1+\tan x}dx=\int_0^{\pi/4}(\ln 2-\ln(1+\tan x))dx=\frac{\pi}{4}\ln 2-I$$</p>
<p>$$2I=\frac{\pi}{4}\ln 2\Rightarrow I=\boxed{\frac{\pi\ln 2}{8}}$$</p>
</div>
Correct Answer: A