Q48 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let $[\cdot]$ be the greatest integer function. If $\alpha=\int_{0}^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi}\left(\frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta}\right) \mathrm{d} \theta$ is equal to $\_\_\_\_$ .
Correct answer: 36
Solution
$\because \int_{0}^{64} x^{\frac{1}{3}} \, dx = \dfrac{3}{4} \left[ x^{\frac{4}{3}} \right]_{0}^{64} = 192$ &
$\int_{0}^{64} [x^{1/3}] \, dx = \int_{0}^{1} [x^{1/3}] \, dx + \int_{1}^{8} [x^{1/3}] \, dx + \int_{8}^{27} [x^{1/3}] \, dx + \int_{27}^{64} [x^{1/3}] \, dx = 156$
So $\alpha = 192 - 156 = 36$
Now $E = \dfrac{1}{\pi} \int_{0}^{36\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$
$= \dfrac{36}{\pi} \int_{0}^{\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$
$\Rightarrow E = \dfrac{36 \cdot 2}{\pi} \int_{0}^{\pi/2} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$
Let $J = \int_{0}^{\pi/2} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$ ....(1)
Applying King
$J = \int_{0}^{\pi/2} \dfrac{\cos^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$ .....(2)
Now $2J = \int_{0}^{\pi/2} \dfrac{1}{\sin^6\theta + \cos^6\theta} \, d\theta$ (add (1) & (2))
$= \int_{0}^{\pi/2} \dfrac{\sec^6\theta}{\tan^6\theta + 1} \, d\theta$
$= \int_{0}^{\infty} \dfrac{(1 + \lambda^2)}{\lambda^4 - \lambda^2 + 1} \, d\lambda$
$= \int_{0}^{\infty} \dfrac{1 + \dfrac{1}{\lambda^2}}{\lambda^2 - 1 + \dfrac{1}{\lambda^2}} \, d\lambda$
$= \pi$
$\Rightarrow J = \dfrac{\pi}{2}$
$\Rightarrow E = \dfrac{36 \cdot 2}{\pi} \times J = 36$
$\int_{0}^{64} [x^{1/3}] \, dx = \int_{0}^{1} [x^{1/3}] \, dx + \int_{1}^{8} [x^{1/3}] \, dx + \int_{8}^{27} [x^{1/3}] \, dx + \int_{27}^{64} [x^{1/3}] \, dx = 156$
So $\alpha = 192 - 156 = 36$
Now $E = \dfrac{1}{\pi} \int_{0}^{36\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$
$= \dfrac{36}{\pi} \int_{0}^{\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$
$\Rightarrow E = \dfrac{36 \cdot 2}{\pi} \int_{0}^{\pi/2} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$
Let $J = \int_{0}^{\pi/2} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$ ....(1)
Applying King
$J = \int_{0}^{\pi/2} \dfrac{\cos^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$ .....(2)
Now $2J = \int_{0}^{\pi/2} \dfrac{1}{\sin^6\theta + \cos^6\theta} \, d\theta$ (add (1) & (2))
$= \int_{0}^{\pi/2} \dfrac{\sec^6\theta}{\tan^6\theta + 1} \, d\theta$
$= \int_{0}^{\infty} \dfrac{(1 + \lambda^2)}{\lambda^4 - \lambda^2 + 1} \, d\lambda$
$= \int_{0}^{\infty} \dfrac{1 + \dfrac{1}{\lambda^2}}{\lambda^2 - 1 + \dfrac{1}{\lambda^2}} \, d\lambda$
$= \pi$
$\Rightarrow J = \dfrac{\pi}{2}$
$\Rightarrow E = \dfrac{36 \cdot 2}{\pi} \times J = 36$
Hence, the correct answer is 36.