If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+C$ where $C$ is the constant of integration, then $\mathrm{f}\left(\frac{\pi}{6}\right)-\mathrm{f}\left(\frac{\pi}{4}\right)$ is equal to

(B) $\frac{4}{\sqrt{3}}(8-\sqrt{6})$

(A) $\frac{1}{\sqrt{3}}(26+\sqrt{3})$

(C) $\frac{1}{\sqrt{3}}(26-\sqrt{3})$

(D) $\frac{2}{\sqrt{3}}(4+\sqrt{6})$

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Indefinite Integrals

\int (\frac{1 - 5 c o s ^{2} x}{s i n ^{5} x c o s ^{2} x}) d x = f (x) + C

where

C

is the constant of integration, then

f (\frac{π}{6}) - f (\frac{π}{4})

is equal to

Answer: (B)

Observing that

\frac{d}{d x} (\frac{t a n x}{s i n ^{5} x}) = \frac{s e c ^{2} x}{s i n ^{5} x} - \frac{5 s i n x}{s i n ^{6} x} = \frac{s e c ^{2} x - 5}{s i n ^{5} x} = \frac{1 - 5 c o s ^{2} x}{s i n ^{5} x c o s ^{2} x}

, the integral is

f (x) = \frac{t a n x}{s i n ^{5} x}

. Then

f (\frac{π}{6}) = \frac{1/ 3}{( 1/2 ) ^{5}} = \frac{32}{3}

and

f (\frac{π}{4}) = \frac{1}{( 1/ 2 ) ^{5}} = 42

. Therefore

f (\frac{π}{6}) - f (\frac{π}{4}) = \frac{32}{3} - 42 = \frac{4}{3} (8 - 6)

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option