JEE Main 2026 Jan 22 Shift 2, Mathematics Q22: Definite Integration And Fundamental Theorem

${\int }_0^{64}{x}^{1/3}dx=\frac{3}{4}{x}^{4/3}{|}_0^{64}=\frac{3}{4}\cdot 256=192$.

${\int }_0^{64}[x^{1/3}]dx=0\cdot 1+1\cdot 7+2\cdot 19+3\cdot 37=0+7+38+111=156$ (using lengths $1^3,2^3-1^3,3^3-2^3,4^3-3^3$).

So $\alpha =192-156=36$.

Let $E=\frac{1}{\pi }{\int }_0^{36\pi }\frac{{\sin }^2\theta }{{\sin }^6\theta +{\cos }^6\theta }d\theta$. The integrand has period $\pi$, so $E=\frac{36}{\pi }{\int }_0^{\pi }\frac{{\sin }^2\theta }{{\sin }^6\theta +{\cos }^6\theta }d\theta =\frac{72}{\pi }{\int }_0^{\pi /2}\frac{{\sin }^2\theta d\theta }{{\sin }^6\theta +{\cos }^6\theta }$.

Let $J={\int }_0^{\pi /2}\frac{{\sin }^2\theta d\theta }{{\sin }^6\theta +{\cos }^6\theta }$. By King's rule, $J$ also equals ${\int }_0^{\pi /2}\frac{{\cos }^2\theta d\theta }{{\sin }^6\theta +{\cos }^6\theta }$.

Adding: $2J={\int }_0^{\pi /2}\frac{d\theta }{{\sin }^6\theta +{\cos }^6\theta }={\int }_0^{\pi /2}\frac{{\sec }^6\theta }{{\tan }^6\theta +1}d\theta$ after dividing by ${\cos }^6\theta$.

Substituting $u=\tan \theta$ and simplifying via the identity ${\tan }^6+1=({\tan }^2+1)({\tan }^4-{\tan }^2+1)$ gives $2J=\pi$, so $J=\frac{\pi }{2}$.

Therefore $E=\frac{72}{\pi }\cdot \frac{\pi }{2}=36$.