Definite IntegrationmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Integral (4 - cosec²x)/cos⁴x = 32/(3√3) | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question

The value of the integral

\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{4-\operatorname{cosec}^2 x}{\cos^4 x}\,dx

\dfrac{11}{\sqrt3}

\dfrac{16}{\sqrt3}

\dfrac{32}{3\sqrt3}

correct

\dfrac{64}{3\sqrt3}

Solution

Step 1: Let

F(x)=\dfrac{\cot x}{\cos^4 x}=\cot x\cdot\sec^4 x

. Then

F'=(-\operatorname{cosec}^2 x)\sec^4 x+\cot x\cdot4\sec^4 x\tan x

; using

\cot x\tan x=1

, the second term is

\dfrac{4}{\cos^4 x}

F'(x)=\frac{4-\operatorname{cosec}^2 x}{\cos^4 x},

\therefore\ \int_{\pi/6}^{\pi/3}\frac{4-\operatorname{cosec}^2 x}{\cos^4 x}\,dx=\left[\frac{\cot x}{\cos^4 x}\right]_{\pi/6}^{\pi/3}.

Step 2: At

x=\dfrac{\pi}{3}

\cot\dfrac{\pi}{3}=\dfrac{1}{\sqrt3}

\cos^4\dfrac{\pi}{3}=\dfrac{1}{16}\Rightarrow\dfrac{1/\sqrt3}{1/16}=\dfrac{16}{\sqrt3}

. At

x=\dfrac{\pi}{6}

\cot\dfrac{\pi}{6}=\sqrt3

\cos^4\dfrac{\pi}{6}=\dfrac{9}{16}\Rightarrow\dfrac{\sqrt3}{9/16}=\dfrac{16\sqrt3}{9}=\dfrac{16}{3\sqrt3}

. Step 3:

\dfrac{16}{\sqrt3}-\dfrac{16}{3\sqrt3}=\dfrac{16}{\sqrt3}\left(1-\dfrac13\right)=\dfrac{16}{\sqrt3}\cdot\dfrac23=\dfrac{32}{3\sqrt3}

. Correct answer: (3)

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