JEE Main 2026 — 5 April, Morning Shift — Question 43

$=\int_{\pi / 6}^{\pi / 3} \frac{4}{\cos ^{4} x} d x-\int_{\pi / 6}^{\pi / 3} \frac{\operatorname{cosec}^{2} x}{\cos ^{4} x} d x$

$$\begin{aligned} & =\int_{\pi / 6}^{\pi / 3} \frac{4}{\cos ^{4} x} d x-\left[-\frac{\cot x}{\cos ^{4} x}-\int_{\pi / 6}^{\pi / 3}(-\cot x) \frac{-4}{\cos ^{5} x}(-\sin x) d x\right] \\& \quad=\left.\frac{\cot x}{\cos ^{4} x}\right|_{\frac{\pi}{6}} ^{\frac{\pi}{3}} \\& \quad=\frac{1 / \sqrt{3}}{(1 / 2)^{4}}-\frac{\sqrt{3}}{\left(\frac{\sqrt{3}}{2}\right)^{4}} \end{aligned}$$