Integral Calculus – Indefinite and Definite Integration – JEE Main 28 Jan 2026 Shift 1

Solution:

Let $I = \int \frac{1 – 5\cos^2 x}{\sin^5 x \cos^2 x} dx$.
Split the numerator:
$$I = \int \frac{1}{\sin^5 x \cos^2 x} dx – \int \frac{5\cos^2 x}{\sin^5 x \cos^2 x} dx$$
$$I = \int \frac{\sec^2 x}{\sin^5 x} dx – 5\int \frac{1}{\sin^5 x} dx$$

Apply Integration by Parts on the first integral.
Let $u = (\sin x)^{-5}$ and $dv = \sec^2 x dx$.
Then $du = -5(\sin x)^{-6}\cos x dx$ and $v = \tan x$.

$$\int \frac{\sec^2 x}{\sin^5 x} dx = \frac{\tan x}{\sin^5 x} – \int \tan x \left(-5 \frac{\cos x}{\sin^6 x}\right) dx$$
$$= \frac{\tan x}{\sin^5 x} + 5 \int \frac{\sin x}{\cos x} \frac{\cos x}{\sin^6 x} dx$$
$$= \frac{\tan x}{\sin^5 x} + 5 \int \frac{1}{\sin^5 x} dx$$

Substitute this back into the expression for $I$:
$$I = \left( \frac{\tan x}{\sin^5 x} + 5 \int cosec^5 x dx \right) – 5 \int cosec^5 x dx$$
The integral terms cancel out:
$$I = \frac{\tan x}{\sin^5 x} + C$$
Thus, $f(x) = \frac{\tan x}{\sin^5 x}$.

Now evaluate $f(\frac{\pi}{6}) – f(\frac{\pi}{4})$:
$$f\left(\frac{\pi}{6}\right) = \frac{\tan 30^{\circ}}{\sin^5 30^{\circ}} = \frac{1/\sqrt{3}}{(1/2)^5} = \frac{1}{\sqrt{3}} \cdot 32 = \frac{32}{\sqrt{3}}$$
$$f\left(\frac{\pi}{4}\right) = \frac{\tan 45^{\circ}}{\sin^5 45^{\circ}} = \frac{1}{(1/\sqrt{2})^5} = (\sqrt{2})^5 = 4\sqrt{2}$$

Difference:
$$f\left(\frac{\pi}{6}\right) – f\left(\frac{\pi}{4}\right) = \frac{32}{\sqrt{3}} – 4\sqrt{2}$$
Taking common denominator $\sqrt{3}$:
$$= \frac{32 – 4\sqrt{6}}{\sqrt{3}} = \frac{4(8 – \sqrt{6})}{\sqrt{3}}$$

Ans. (2)