JEE Main 2026 — 22 January, Evening Shift — Question 22

$\because \int_{0}^{64} \mathrm{x}^{\frac{1}{3}} \mathrm{dx}=\frac{3}{4} \cdot\left[\mathrm{x}^{\frac{4}{3}}\right]_{0}^{64}=192 \&$

$\int_{0}^{64}\left[x^{1 / 3}\right] d x=\int_{0}^{1}\left[x^{1 / 3}\right] d x+\int_{1}^{8}\left[x^{1 / 3}\right] d x+\int_{8}^{27}\left[x^{1 / 3}\right] d x+\int_{27}^{64}\left[x^{1 / 3}\right] d x=156$

So $\alpha=192-156=36$

Now $E=\frac{1}{\pi} \int_{0}^{36 \pi} \frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta} d \theta$

$=\frac{36}{\pi} \int_{0}^{\pi} \frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta} d \theta$

$\Rightarrow \mathrm{E}=\frac{36 \cdot 2}{\pi} \int_{0}^{\pi / 2} \frac{\sin ^{2} \theta \mathrm{~d} \theta}{\sin ^{6} \theta+\cos ^{6} \theta}$

Let $\mathrm{J}=\int_{0}^{\pi / 2} \frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta} \mathrm{~d} \theta$

Applying King $\begin{gathered} J=\int_{0}^{\pi / 2} \frac{\cos ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta} d \theta \end{gathered}$

Now $2 \mathrm{~J}=\int_{0}^{\pi / 2} \frac{1}{\sin ^{6} \theta+\cos ^{6} \theta} \mathrm{~d} \theta$

$=\int_{0}^{\pi / 2} \frac{\sec ^{6} \theta}{\tan ^{6} \theta+1} \mathrm{~d} \theta$

$=\int_{0}^{\infty} \frac{\left(1+\lambda^{2}\right)}{\lambda^{4}-\lambda^{2}+1} \mathrm{~d} \lambda$

$=\int_{0}^{\infty} \frac{1+\frac{1}{\lambda^{2}}}{\lambda^{2}-1+\frac{1}{\lambda^{2}}} d \lambda$ $=\pi$

$\Rightarrow \mathrm{J}=\frac{\pi}{2}$

$\Rightarrow \mathrm{E}=\frac{36 \cdot 2}{\pi} \times \mathrm{J}=36$