Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/4}\log(1+\tan x)\,dx$

The key idea is to use the substitution $x \to \frac{\pi}{4} - x$, which exploits the symmetry of the integrand under the transformation $\tan x \to \frac{1 - \tan x}{1 + \tan x}$.

Let $I = \int_{0}^{\pi/4} \log(1 + \tan x) \, dx$.

Substitute $x = \frac{\pi}{4} - t$, so $dx = -dt$. When $x = 0$, $t = \frac{\pi}{4}$; when $x = \frac{\pi}{4}$, $t = 0$. Then:

Simplify inside the log: $1 + \frac{1 - \tan t}{1 + \tan t} = \frac{2}{1 + \tan t}$. Thus: …

The key insight here is that the integrand $\log(1+\tan x)$ doesn't have a simple antiderivative, but the limits $0$ to $\pi/4$ are special. When you replace $x$ by $\pi/4 - x$, the $\tan$ function transforms in a neat way, and the sum of the original and transformed integrals collapses to a constant times the length of the interval.

Let's work through it.

1. Set up the transformation. Let $I = \int_{0}^{\pi/4} \log(1+\tan x)\,dx$. Use the property $\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx$ with $a = \pi/4$. So

2. Simplify $\tan(\pi/4 - x)$. Recall the identity:

This is a standard result from $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ with $A = \pi/4$, $\tan A = 1$.

3. Rewrite the transformed integrand. Substitute into the log:

Therefore

4. Add the two expressions for $I$. We now have:

Adding them: …

Method: Self-referential reflection — when $f(a-x)=C-f(x)$

If reflecting the integrand reproduces a constant minus the original (a self-referential relation), adding the two forms lets you solve for $I$ algebraically.

Steps

Step 1: Reflect with $x\to a-x$.

Apply $I=\displaystyle\int_0^a f(a-x)\,dx$ and simplify using the relevant identity, e.g. $\tan(\tfrac{\pi}{4}-x)=\dfrac{1-\tan x}{1+\tan x}$.

Step 2: Expose the constant-minus-original structure. …

Common Mistakes

Mistake 1: Misapplying the tangent-difference identity.

Why it's wrong: $\tan(\tfrac{\pi}{4}-x)=\tfrac{1-\tan x}{1+\tan x}$; getting the numerator/denominator swapped breaks the whole reduction. Correct approach: use $\tan(A-B)$ with $\tan\tfrac{\pi}{4}=1$.

Mistake 2: Not simplifying $1+\tfrac{1-\tan x}{1+\tan x}$ to $\tfrac{2}{1+\tan x}$. …