Q.Evaluate $\int_{\pi/6}^{\pi/3} \dfrac{dx}{1 + \sqrt{\tan x}}$

Concept: Definite Integral Symmetry (King’s property)

Use the substitution $x \to a + b - x$, where $a = \pi/6$, $b = \pi/3$.

Let $I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}}$.

Step 1: Replace $x$ by $\pi/2 - x$ (since $a+b = \pi/2$).

Then $dx \to -dx$, and the limits swap, giving:

Step 2: Since $\sqrt{\cot x} = 1/\sqrt{\tan x}$, rewrite:

$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \frac{1}{\sqrt{\tan x}}} = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x} \, dx}{1 + \sqrt{\tan x}}.$ …

When you see an integral with a complicated function like $\sqrt{\tan x}$, the first instinct might be to try a substitution. But here, the limits $\pi/6$ and $\pi/3$ are symmetric about $\pi/4$, and the integrand has a special structure. The key is to use a symmetry property of definite integrals — often called the "King's property" — which lets you replace $x$ with $a+b-x$ without changing the value of the integral. This trick is especially powerful when the integrand has terms like $\tan x$ and $\cot x$ that swap under this transformation.

Let’s see how it works.

1. State the property. For any continuous function $f$ on $[a, b]$, we have:

Here, $a = \pi/6$ and $b = \pi/3$, so $a+b = \pi/2$. Thus:

2. Simplify the transformed integrand. Recall that $\tan(\pi/2 - x) = \cot x = \frac{1}{\tan x}$. So:

Therefore:

3. Add the two expressions for $I$. We now have two forms of the same integral:

Adding them: …

Method: The reflection (a+b-x) property for a self-complementary integrand

Use this when a definite integral $\int_a^b f(x)\,dx$ has an integrand that turns into "$1$ minus itself" under the substitution $x \to a+b-x$ — typically a fraction of the form $\dfrac{1}{1+g(x)}$ where $g(a+b-x)=\dfrac{1}{g(x)}$.

Steps

Step 1: Write down the reflection property.

For any continuous $f$ on $[a,b]$,

This does not change the value — it only re-expresses the same area, reading the interval from the other end.

Step 2: Compute $a+b$ and substitute.

Add the two endpoints to find the "reflection centre" $a+b$. Replace $x$ by $a+b-x$ inside $f$ and simplify each trig/algebraic piece (e.g. $\tan(\tfrac{\pi}{2}-x)=\cot x$, so $\sqrt{\tan}$ becomes $\sqrt{\cot}=1/\sqrt{\tan}$). …

Common Mistakes

Mistake 1: Trying to integrate $\dfrac{1}{1+\sqrt{\tan x}}$ directly.

Why it's wrong: this integrand has no elementary antiderivative you could find in an exam, so a $t=\sqrt{\tan x}$ substitution just produces an unmanageable rational function. Correct approach: recognise the symmetric limits and apply the $x\to a+b-x$ reflection property instead.

Mistake 2: Using the wrong reflection centre. …