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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Definite Integration

Let [·] be the greatest integer function. If α=∫_ 0 ^ 64 (x^ 1 / 3...

Let [⋅][\cdot] be the greatest integer function. If α=∫064(x1/3−[x1/3])dx\alpha=\int_{0}^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x, then 1π∫0απ(sin⁡2θsin⁡6θ+cos⁡6θ)dθ\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

0

Step-by-step solution

∵∫064x13 dx=34[x43]064=192\because \int_{0}^{64} x^{\frac{1}{3}} \, dx = \dfrac{3}{4} \left[ x^{\frac{4}{3}} \right]_{0}^{64} = 192 &
∫064[x1/3] dx=∫01[x1/3] dx+∫18[x1/3] dx+∫827[x1/3] dx+∫2764[x1/3] dx=156\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 α=192−156=36\alpha = 192 - 156 = 36
Now E=1π∫036πsin⁡2θsin⁡6θ+cos⁡6θ dθE = \dfrac{1}{\pi} \int_{0}^{36\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta
=36π∫0πsin⁡2θsin⁡6θ+cos⁡6θ dθ= \dfrac{36}{\pi} \int_{0}^{\pi} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta
⇒E=36⋅2π∫0π/2sin⁡2θsin⁡6θ+cos⁡6θ dθ\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=∫0π/2sin⁡2θsin⁡6θ+cos⁡6θ dθJ = \int_{0}^{\pi/2} \dfrac{\sin^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta ....(1)
Applying King
J=∫0π/2cos⁡2θsin⁡6θ+cos⁡6θ dθJ = \int_{0}^{\pi/2} \dfrac{\cos^2\theta}{\sin^6\theta + \cos^6\theta} \, d\theta .....(2)
Now 2J=∫0π/21sin⁡6θ+cos⁡6θ dθ2J = \int_{0}^{\pi/2} \dfrac{1}{\sin^6\theta + \cos^6\theta} \, d\theta (add (1) & (2))
=∫0π/2sec⁡6θtan⁡6θ+1 dθ= \int_{0}^{\pi/2} \dfrac{\sec^6\theta}{\tan^6\theta + 1} \, d\theta
=∫0∞(1+λ2)λ4−λ2+1 dλ= \int_{0}^{\infty} \dfrac{(1 + \lambda^2)}{\lambda^4 - \lambda^2 + 1} \, d\lambda
=∫0∞1+1λ2λ2−1+1λ2 dλ= \int_{0}^{\infty} \dfrac{1 + \dfrac{1}{\lambda^2}}{\lambda^2 - 1 + \dfrac{1}{\lambda^2}} \, d\lambda
=π= \pi
⇒J=π2\Rightarrow J = \dfrac{\pi}{2}
⇒E=36⋅2π×J=36\Rightarrow E = \dfrac{36 \cdot 2}{\pi} \times J = 36
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