If ∫( 1-5 ^ 2 x ^ 5 x ^ 2 x ) d x=f(x)+C , where C is the constant of integration, then f((π)/(6))-f((π)/(4)) is equal to

Correct answer: C. 4 3 (8- 6 ) ∫ (1 - 5 ^2 x)/( ^5 x ^2 x) dx = ∫ (dx)/( ^5 x ^2 x) - 5∫ (dx)/( ^5 x) = ∫ ( ^2 x \, dx)/( ^5 x) - 5∫ (dx)/( ^5 x) By IBP: = ( x)/( ^5 x) - ∫ (-(5)/( ^6 x)) x · x \, dx - 5∫ (dx)/( ^5 x) = ( x)/( ^5 x) + C So f(x) = ( x)/( ^5 x) f((π)/(6)) - f((π)/(4)) = 2^5 3 - ( 2 )^5 = 32 3 - 4 2 = 4 3 (8 - 6 )

JEE Main 202628 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Definite Integration

If ∫( 1-5 ^ 2 x ^ 5 x ^ 2 x ) d x=f(x)+C , where C is the constant...

\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

Options

A $\frac{1}{\sqrt{3}}(26+\sqrt{3})$
B $\frac{1}{\sqrt{3}}(26-\sqrt{3})$
C $\frac{4}{\sqrt{3}}(8-\sqrt{6})$
D $\frac{2}{\sqrt{3}}(4+\sqrt{6})$

Correct answer

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

Step-by-step 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

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})

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