Q41 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to
- A$\frac{1}{\sqrt{3}}(26+\sqrt{3})$
- B$\frac{1}{\sqrt{3}}(26-\sqrt{3})$
- Correct answerC$\frac{4}{\sqrt{3}}(8-\sqrt{6})$
- D$\frac{2}{\sqrt{3}}(4+\sqrt{6})$
Solution
$\int \frac{1 - 5\cos^2 x}{\sin^5 x \cos^2 x} dx = \int \frac{dx}{\sin^5 x \cos^2 x} - 5\int \frac{dx}{\sin^5 x}$
$= \int \frac{\sec^2 x \, dx}{\sin^5 x} - 5\int \frac{dx}{\sin^5 x}$
By IBP:
$= \frac{\tan x}{\sin^5 x} - \int \left(-\frac{5}{\sin^6 x}\right) \cos x \cdot \tan x \, dx - 5\int \frac{dx}{\sin^5 x}$
$= \frac{\tan x}{\sin^5 x} + C$
So $f(x) = \frac{\tan x}{\sin^5 x}$
$f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right) = \frac{2^5}{\sqrt{3}} - (\sqrt{2})^5 = \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{4}{\sqrt{3}}(8 - \sqrt{6})$
$= \int \frac{\sec^2 x \, dx}{\sin^5 x} - 5\int \frac{dx}{\sin^5 x}$
By IBP:
$= \frac{\tan x}{\sin^5 x} - \int \left(-\frac{5}{\sin^6 x}\right) \cos x \cdot \tan x \, dx - 5\int \frac{dx}{\sin^5 x}$
$= \frac{\tan x}{\sin^5 x} + C$
So $f(x) = \frac{\tan x}{\sin^5 x}$
$f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right) = \frac{2^5}{\sqrt{3}} - (\sqrt{2})^5 = \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{4}{\sqrt{3}}(8 - \sqrt{6})$
Hence, the correct answer is option: 'C'.