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 ∫π/6π/34−cosec⁡2xcos⁡4x dx\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{4-\operatorname{cosec}^2 x}{\cos^4 x}\,dx is
A113\dfrac{11}{\sqrt3}
B163\dfrac{16}{\sqrt3}
C3233\dfrac{32}{3\sqrt3}correct
D6433\dfrac{64}{3\sqrt3}
Solution
Step 1: Let F(x)=cot⁡xcos⁡4x=cot⁡x⋅sec⁡4xF(x)=\dfrac{\cot x}{\cos^4 x}=\cot x\cdot\sec^4 x. Then F′=(−cosec⁡2x)sec⁡4x+cot⁡x⋅4sec⁡4xtan⁡xF'=(-\operatorname{cosec}^2 x)\sec^4 x+\cot x\cdot4\sec^4 x\tan x; using cot⁡xtan⁡x=1\cot x\tan x=1, the second term is 4cos⁡4x\dfrac{4}{\cos^4 x}:
F′(x)=4−cosec⁡2xcos⁡4x,F'(x)=\frac{4-\operatorname{cosec}^2 x}{\cos^4 x},
∴ ∫π/6π/34−cosec⁡2xcos⁡4x dx=[cot⁡xcos⁡4x]π/6π/3.\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=π3x=\dfrac{\pi}{3}: cot⁡π3=13\cot\dfrac{\pi}{3}=\dfrac{1}{\sqrt3}, cos⁡4π3=116⇒1/31/16=163\cos^4\dfrac{\pi}{3}=\dfrac{1}{16}\Rightarrow\dfrac{1/\sqrt3}{1/16}=\dfrac{16}{\sqrt3}. At x=π6x=\dfrac{\pi}{6}: cot⁡π6=3\cot\dfrac{\pi}{6}=\sqrt3, cos⁡4π6=916⇒39/16=1639=1633\cos^4\dfrac{\pi}{6}=\dfrac{9}{16}\Rightarrow\dfrac{\sqrt3}{9/16}=\dfrac{16\sqrt3}{9}=\dfrac{16}{3\sqrt3}. Step 3: 163−1633=163(1−13)=163⋅23=3233\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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