Integral Calculus – Definite Integration – JEE Main 23 Jan 2026 Shift 1

Solution:

Let $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+(\tan 2x)^{1/4}}$.

Using King’s Property: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.
Here, $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$.

Replace $x$ with $(\frac{\pi}{4}-x)$. The term $\tan(2x)$ becomes:
$$ \tan(2(\frac{\pi}{4}-x)) = \tan(\frac{\pi}{2}-2x) = \cot(2x) = \frac{1}{\tan 2x} $$

The integral becomes:
$$ I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+(\cot 2x)^{1/4}} $$
$$ I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\frac{1}{(\tan 2x)^{1/4}}} $$
$$ I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{(\tan 2x)^{1/4}}{(\tan 2x)^{1/4}+1} dx $$

Adding the two expressions for $I$:
$$ 2I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \left( \frac{1}{1+(\tan 2x)^{1/4}} + \frac{(\tan 2x)^{1/4}}{1+(\tan 2x)^{1/4}} \right) dx $$
$$ 2I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} 1 dx $$

$$ 2I = [x]_{\frac{\pi}{24}}^{\frac{5\pi}{24}} = \frac{5\pi}{24} – \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6} $$

$$ I = \frac{\pi}{12} $$

Ans. (1)