JEE Main 2026 Apr 5 Shift 1, Mathematics Q20: Definite Integration And Fundamental Theorem

Observe that $\frac{d}{dx}\left(\frac{\cot x}{{\cos }^{4}x}\right)=\frac{-{\csc }^{2}x}{{\cos }^{4}x}+\cot x\cdot \frac{4\sin x}{{\cos }^{5}x}=\frac{-{\csc }^{2}x+4}{{\cos }^{4}x}$.

So the integrand is exactly $\frac{d}{dx}\left(\frac{\cot x}{{\cos }^{4}x}\right)$, and the integral equals ${\left[\frac{\cot x}{{\cos }^{4}x}\right]}_{\pi /6}^{\pi /3}$.

At $x=\pi /3$: $\frac{1/\sqrt{3}}{(1/2{)}^4}=\frac{16}{\sqrt{3}}$. At $x=\pi /6$: $\frac{\sqrt{3}}{(\sqrt{3}/2{)}^4}=\frac{\sqrt{3}\cdot 16}{9}=\frac{16}{3\sqrt{3}}$.

Therefore the integral $=\frac{16}{\sqrt{3}}-\frac{16}{3\sqrt{3}}=\frac{16}{\sqrt{3}}\left(1-\frac{1}{3}\right)=\frac{32}{3\sqrt{3}}$.