JEE Main 2026 — Mathematics Calculus

The given integral is $\displaystyle I = \displaystyle\int_{\pi/6}^{\pi/3} \left(\dfrac{4 - \csc^2 x}{\cos^4 x}\right) dx$I=∫π/6π/3​(cos4x4−csc2x​)dx.

We can rewrite the integrand in terms of $\tan x$tanx and $\sec x$secx. Using the identity $\csc^2 x = 1 + \cot^2 x = 1 + \dfrac{1}{\tan^2 x}$csc2x=1+cot2x=1+tan2x1​, we get:

$$I = \displaystyle\int_{\pi/6}^{\pi/3} \left(4 - 1 - \dfrac{1}{\tan^2 x}\right) \sec^4 x \, dx$$I=∫π/6π/3​(4−1−tan2x1​)sec4xdx

$$I = \displaystyle\int_{\pi/6}^{\pi/3} \left(3 - \dfrac{1}{\tan^2 x}\right) (1 + \tan^2 x) \sec^2 x \, dx$$I=∫π/6π/3​(3−tan2x1​)(1+tan2x)sec2xdx

Substitute $t = \tan x$t=tanx, which gives $dt = \sec^2 x \, dx$dt=sec2xdx.

The limits of integration change as follows:

When $x = \dfrac{\pi}{6}$x=6π​, $t = \dfrac{1}{\sqrt{3}}$t=3​1​.

When $x = \dfrac{\pi}{3}$x=3π​, $t = \sqrt{3}$t=3​.

The integral becomes:

$$I = \displaystyle\int_{1/\sqrt{3}}^{\sqrt{3}} \left(3 - \dfrac{1}{t^2}\right) (1 + t^2) dt$$I=∫1/3​3​​(3−t21​)(1+t2)dt

Expanding the integrand, we get:

$$I = \displaystyle\int_{1/\sqrt{3}}^{\sqrt{3}} \left(3 + 3t^2 - \dfrac{1}{t^2} - 1\right) dt$$I=∫1/3​3​​(3+3t2−t21​−1)dt

$$I = \displaystyle\int_{1/\sqrt{3}}^{\sqrt{3}} \left(3t^2 + 2 - t^{-2}\right) dt$$I=∫1/3​3​​(3t2+2−t−2)dt

Integrating with respect to $t$t:

$$I = \left[ t^3 + 2t + \dfrac{1}{t} \right]_{1/\sqrt{3}}^{\sqrt{3}}$$I=[t3+2t+t1​]1/3​3​​

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

$$\left( \sqrt{3} \right)^3 + 2\sqrt{3} + \dfrac{1}{\sqrt{3}} = 3\sqrt{3} + 2\sqrt{3} + \dfrac{\sqrt{3}}{3} = \dfrac{16\sqrt{3}}{3} = \dfrac{48}{3\sqrt{3}}$$(3​)3+23​+3​1​=33​+23​+33​​=3163​​=33​48​

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

$$\left( \dfrac{1}{\sqrt{3}} \right)^3 + 2\left( \dfrac{1}{\sqrt{3}} \right) + \sqrt{3} = \dfrac{1}{3\sqrt{3}} + \dfrac{2}{\sqrt{3}} + \sqrt{3} = \dfrac{1 + 6 + 9}{3\sqrt{3}} = \dfrac{16}{3\sqrt{3}}$$(3​1​)3+2(3​1​)+3​=33​1​+3​2​+3​=33​1+6+9​=33​16​

Subtracting the lower limit value from the upper limit value:

$$I = \dfrac{48}{3\sqrt{3}} - \dfrac{16}{3\sqrt{3}} = \dfrac{32}{3\sqrt{3}}$$I=33​48​−33​16​=33​32​

Answer: $\dfrac{32}{3\sqrt{3}}$33​32​