JEE Main 2026 — Mathematics Calculus

$\displaystyle \because \int_{0}^{64} x^{\frac{1}{3}} \, dx = \dfrac{3}{4} \left[ x^{\frac{4}{3}} \right]_{0}^{64} = 192$∵∫064​x31​dx=43​[x34​]064​=192 &

$$\int_{0}^{64} [x^{1/3}] \, dx = \int_{0}^{1} [x^{1/3}] \, dx + \int_{1}^{8} [x^{1/3}] \, dx + \int_{8}^{27} [x^{1/3}] \, dx + \int_{27}^{64} [x^{1/3}] \, dx = 156$$∫064​[x1/3]dx=∫01​[x1/3]dx+∫18​[x1/3]dx+∫827​[x1/3]dx+∫2764​[x1/3]dx=156

So $\alpha = 192 - 156 = 36$α=192−156=36

Now $\displaystyle E = \dfrac{1}{\pi} \int_{0}^{36\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$E=π1​∫036π​sin6θ+cos6θsin2θ​dθ

$$= \dfrac{36}{\pi} \int_{0}^{\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$$=π36​∫0π​sin6θ+cos6θsin2θ​dθ

$$\Rightarrow E = \dfrac{36 \cdot 2}{\pi} \int_{0}^{\pi/2} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$$⇒E=π36⋅2​∫0π/2​sin6θ+cos6θsin2θ​dθ

Let $\displaystyle J = \int_{0}^{\pi/2} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$J=∫0π/2​sin6θ+cos6θsin2θ​dθ ....(1)

Applying King

$\displaystyle J = \int_{0}^{\pi/2} \dfrac{\cos^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta$J=∫0π/2​sin6θ+cos6θcos2θ​dθ .....(2)

Now $\displaystyle 2J = \int_{0}^{\pi/2} \dfrac{1}{\sin^6\theta + \cos^6\theta} \, d\theta$2J=∫0π/2​sin6θ+cos6θ1​dθ (add (1) & (2))

$$= \int_{0}^{\pi/2} \dfrac{\sec^6\theta}{\tan^6\theta + 1} \, d\theta$$=∫0π/2​tan6θ+1sec6θ​dθ

$$= \int_{0}^{\infty} \dfrac{(1 + \lambda^2)}{\lambda^4 - \lambda^2 + 1} \, d\lambda$$=∫0∞​λ4−λ2+1(1+λ2)​dλ

$$= \int_{0}^{\infty} \dfrac{1 + \dfrac{1}{\lambda^2}}{\lambda^2 - 1 + \dfrac{1}{\lambda^2}} \, d\lambda$$=∫0∞​λ2−1+λ21​1+λ21​​dλ

$$= \pi$$=π

$$\Rightarrow J = \dfrac{\pi}{2}$$⇒J=2π​

$$\Rightarrow E = \dfrac{36 \cdot 2}{\pi} \times J = 36$$⇒E=π36⋅2​×J=36