JEE Main 2026 — Definite Integration Question with Solution

The given integral is $I={\int }_{\pi /6}^{\pi /3}\left(\frac{4-{\csc }^{2}x}{{\cos }^{4}x}\right)dx$.

We can rewrite the integrand in terms of $\tan x$ and $\sec x$. Using the identity ${\csc }^{2}x=1+{\cot }^{2}x=1+\frac{1}{{\tan }^{2}x}$, we get:

$I={\int }_{\pi /6}^{\pi /3}\left(4-1-\frac{1}{{\tan }^{2}x}\right){\sec }^{4}xdx$

$I={\int }_{\pi /6}^{\pi /3}\left(3-\frac{1}{{\tan }^{2}x}\right)(1+{\tan }^{2}x){\sec }^{2}xdx$

Substitute $t=\tan x$, which gives $dt={\sec }^{2}xdx$.

The limits of integration change as follows:
When $x=\frac{\pi }{6}$, $t=\frac{1}{\sqrt{3}}$.
When $x=\frac{\pi }{3}$, $t=\sqrt{3}$.

The integral becomes:

$I={\int }_{1/\sqrt{3}}^{\sqrt{3}}\left(3-\frac{1}{t^2}\right)(1+t^2)dt$

Expanding the integrand, we get:

$I={\int }_{1/\sqrt{3}}^{\sqrt{3}}\left(3+3t^2-\frac{1}{t^2}-1\right)dt$

$I={\int }_{1/\sqrt{3}}^{\sqrt{3}}\left(3t^2+2-t^{-2}\right)dt$

Integrating with respect to $t$:

$I={\left[t^3+2t+\frac{1}{t}\right]}_{1/\sqrt{3}}^{\sqrt{3}}$

Substitute the upper limit $t=\sqrt{3}$:

${\left(\sqrt{3}\right)}^3+2\sqrt{3}+\frac{1}{\sqrt{3}}=3\sqrt{3}+2\sqrt{3}+\frac{\sqrt{3}}{3}=\frac{16\sqrt{3}}{3}=\frac{48}{3\sqrt{3}}$

Substitute the lower limit $t=\frac{1}{\sqrt{3}}$:

${\left(\frac{1}{\sqrt{3}}\right)}^3+2\left(\frac{1}{\sqrt{3}}\right)+\sqrt{3}=\frac{1}{3\sqrt{3}}+\frac{2}{\sqrt{3}}+\sqrt{3}=\frac{1+6+9}{3\sqrt{3}}=\frac{16}{3\sqrt{3}}$

Subtracting the lower limit value from the upper limit value:

$I=\frac{48}{3\sqrt{3}}-\frac{16}{3\sqrt{3}}=\frac{32}{3\sqrt{3}}$

Answer: $\frac{32}{3\sqrt{3}}$