Let \([\bullet]\) be the greatest integer function. If \(\alpha=\int_{0}^{64}\left(\mathrm{x}^{1 / 3}-\left[\mathrm{x}^{1 / 3}\right]\right) \mathrm{dx}\), 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 ______.

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Definite Integrals

Let

[∙]

be the greatest integer function. If

α = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx

, then

\frac{1}{π} \int_{0}^{α π} (\frac{s i n ^{2} θ}{s i n ^{6} θ + c o s ^{6} θ}) d θ

is equal to ______.

Answer: 36

First,

\int_{0}^{64} x^{1/3} d x = \frac{3}{4} \cdot 6 4^{4/3} = 192

and

\int_{0}^{64} [x^{1/3}] d x = 1 \cdot 7 + 2 \cdot 19 + 3 \cdot 37 = 156

, so

α = 192 - 156 = 36

. Let

I = \int_{0}^{π /2} \frac{sin ^{2} θ}{sin ^{6} θ + cos ^{6} θ} d θ

. The substitution

θ \to \frac{π}{2} - θ

gives the same integral with

cos^{2} θ

in the numerator, so

2 I = \int_{0}^{π /2} \frac{d θ}{sin ^{6} θ + cos ^{6} θ} = \int_{0}^{π /2} \frac{d θ}{1 - \frac{3}{4} sin ^{2} 2 θ} = π,

hence

I = \frac{π}{2}

. As the integrand has period

π

\frac{1}{π} \int_{0}^{36 π} \frac{sin ^{2} θ}{sin ^{6} θ + cos ^{6} θ} d θ = \frac{36}{π} \cdot 2 I = \frac{36}{π} \cdot π = 36.

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer