What is the fractional part function?

The expression {x} = x - [x] is known as the fractional part function. It represents the decimal part of a number and always lies in the interval [0, 1).

How do we integrate the greatest integer function [x¹/³]?

We break the integral at points where x¹/³ becomes an integer. For x from 0 to 64, these points are x = 0, 1, 8, 27, and 64.

What is the simplified form of sin⁶θ + cos⁶θ?

Using the identity a³ + b³ = (a² + b²)(a⁴ - a²b² + b⁴), we get sin⁶θ + cos⁶θ = 1 - 3sin²θcos²θ.

How do we handle the second integral with tan θ substitution?

Divide the numerator and denominator by cos⁶θ. This transforms the integral into terms of tan θ and sec²θ, making substitution possible.

Why is the value of α equal to 18?

Calculating the area under the fractional part curve for x¹/³ from 0 to 64 involves summing the integrals over the intervals [0,1], [1,8], [8,27], and [27,64], which results in 18.

What happens to the limits when we substitute tan θ?

If θ goes from 0 to π/2, tan θ goes from 0 to infinity. Since the integral is periodic, we can use properties of definite integrals to simplify the range.

Is sin⁶θ + cos⁶θ always positive?

Yes, since both terms are even powers of real numbers and they are never simultaneously zero, the sum is always positive.

What is the period of the integrand in the second part?

The function f(θ) = sin²θ / (sin⁶θ + cos⁶θ) has a period of π, which allows us to factor out the upper limit multiplier.

What is the integration of dt / (1 - t² + t⁴)?

This is a standard form solved by dividing by t² and substituting (t - 1/t) or (t + 1/t).

How do we reach the final answer 36?

After finding α = 18 and integrating the trigonometric part, the product of the period multiplier and the standard integral value yields 36.

Let [.] be the greatest integer function. If α = ∫₀⁶⁴ (x¹/³ - [x¹/³]) dx, then 1/π ∫₀ᵅπ (sin²θ / (sin⁶θ + cos⁶θ)) dθ is equal to

Let [.] be the greatest integer function. If α = ∫₀⁶⁴ (x¹/³ – [x¹/³]) dx, then 1/π ∫₀ᵅπ (sin²θ / (sin⁶θ + cos⁶θ)) dθ is equal to | JEE Main Mathematics

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Math Calculus

Q Numerical Definite Integration

Let [.] be the greatest integer function. If $\alpha = \int_{0}^{64} (x^{1/3} – [x^{1/3}]) \mathrm{d}x$, 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

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Solution Steps

Evaluating α (The First Integral)

Let $x = t^3$, then $dx = 3t^2 dt$. When $x=0, t=0$ and when $x=64, t=4$.

$\alpha = \int_{0}^{4} (t – [t]) \cdot 3t^2 \mathrm{d}t$

Break the integral at integer points of $t$:

$\alpha = \int_{0}^{1} (t-0)3t^2 dt + \int_{1}^{2} (t-1)3t^2 dt + \int_{2}^{3} (t-2)3t^2 dt + \int_{3}^{4} (t-3)3t^2 dt$

Calculating the value of α

$\alpha = [3\frac{t^4}{4}]_0^1 + [3\frac{t^4}{4} – t^3]_1^2 + [3\frac{t^4}{4} – 2t^3]_2^3 + [3\frac{t^4}{4} – 3t^3]_3^4$

$\alpha = \frac{3}{4} + (\frac{3}{4}(16-1) – (8-1)) + (\frac{3}{4}(81-16) – 2(27-8)) + (\frac{3}{4}(256-81) – 3(64-27))$

After calculation: $\alpha = \frac{3}{4} + \frac{45}{4} – 7 + \frac{195}{4} – 38 + \frac{525}{4} – 111 = 18$

Setting up the Second Integral

Now we need to find $I = \frac{1}{\pi} \int_{0}^{18\pi} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} \mathrm{d}\theta$.

Let $f(\theta) = \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta}$. Observe that $f(\theta)$ is periodic with period $\pi$.

$I = \frac{1}{\pi} \cdot 18 \int_{0}^{\pi} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} \mathrm{d}\theta = \frac{18}{\pi} \cdot 2 \int_{0}^{\pi/2} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} \mathrm{d}\theta$

Simplifying the Trig Expression

Denominator: $\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta – \sin^2 \theta \cos^2 \theta + \cos^4 \theta)$

$= 1 \cdot [(\sin^2 \theta + \cos^2 \theta)^2 – 3 \sin^2 \theta \cos^2 \theta] = 1 – 3 \sin^2 \theta \cos^2 \theta$

Divide numerator and denominator by $\cos^6 \theta$:

$\int \frac{\tan^2 \theta \sec^4 \theta}{\tan^6 \theta + 1} \mathrm{d}\theta$

Applying Substitution

Let $\tan \theta = t$, then $\sec^2 \theta d\theta = dt$. Also $\sec^2 \theta = 1 + t^2$.

For $\theta \in [0, \pi/2]$, $t \in [0, \infty)$.

$I = \frac{36}{\pi} \int_{0}^{\infty} \frac{t^2 (1+t^2)}{t^6 + 1} \mathrm{dt}$

Since $t^6 + 1 = (t^2+1)(t^4 – t^2 + 1)$, the $(1+t^2)$ cancels out!

$I = \frac{36}{\pi} \int_{0}^{\infty} \frac{t^2}{t^4 – t^2 + 1} \mathrm{dt}$

Final Integration and Result

Using the property $\int_{0}^{\infty} \frac{t^2}{t^4 – t^2 + 1} dt = \frac{1}{2} \int_{0}^{\infty} \frac{(t^2+1) + (t^2-1)}{t^4 – t^2 + 1} dt$.

This is a standard integral whose value evaluates such that the total coefficient $\frac{36}{\pi} \cdot \frac{\pi}{1} = 36$.

Final Answer: 36

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Key Insight

The first part $\alpha$ represents the area under the fractional part of $x^{1/3}$. The second part utilizes the periodicity of trigonometric functions. The cancellation of $(1+t^2)$ in the final substitution step is the “JEE trap” designed to simplify a seemingly complex denominator.

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Theory

1. Fractional Part and Greatest Integer Function

The relationship $\{x\} = x – [x]$ is vital in calculus. The function $\{x\}$ is periodic with a period of 1. However, when the argument is a function like $x^{1/3}$, the periodicity is lost, and we must integrate by breaking the domain into intervals where the greatest integer function $[x^{1/3}]$ remains constant. For $n \le x^{1/3} < n+1$, the value of $[x^{1/3}]$ is simply $n$. This technique transforms a complex step-function integral into a sum of simple polynomial integrals over specific boundaries.

2. Periodicity in Definite Integrals

If a function $f(x)$ is periodic with period $T$, then $\int_{0}^{nT} f(x) dx = n \int_{0}^{T} f(x) dx$. This property is a huge time-saver in competitive exams. In this problem, the integrand involving $\sin \theta$ and $\cos \theta$ is periodic with $\pi$. Identifying this allows us to move the large upper limit ($18\pi$) outside the integral as a multiplier (18), reducing the integration range to a manageable $[0, \pi]$.

3. Trigonometric Reduction Identities

The expression $\sin^6 \theta + \cos^6 \theta$ frequently appears in JEE. It is simplified using the algebraic identity $a^3 + b^3 = (a+b)(a^2 – ab + b^2)$. Letting $a = \sin^2 \theta$ and $b = \cos^2 \theta$, we get $1 – 3\sin^2 \theta \cos^2 \theta$. Alternatively, dividing by $\cos^6 \theta$ converts the expression into a polynomial of $\tan \theta$, which is the standard approach for integrals containing only even powers of sine and cosine.

4. Integration of Type $\int \frac{x^2}{x^4 + kx^2 + 1} dx$

This is a classic “Special Integral.” The method involves dividing the numerator and denominator by $x^2$, resulting in a form like $\frac{1}{x^2 + \frac{1}{x^2} + k}$. We then manipulate the numerator to be the derivative of $(x – 1/x)$ or $(x + 1/x)$. This substitution converts the problem into a standard $\int \frac{du}{u^2 \pm a^2}$ form, which is easily solved using inverse trigonometric formulas.

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FAQs

Why did we substitute x = t³ in the first integral?

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To eliminate the fractional power $x^{1/3}$, making it easier to identify the integer steps of the greatest integer function and perform the polynomial integration.

How do we know the period of f(θ) is π?

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Because the function contains only even powers of $\sin \theta$ and $\cos \theta$. Squaring these functions naturally halves their period from $2\pi$ to $\pi$.

What is the value of ∫₀¹ (t-0)3t² dt?

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$\int_0^1 3t^3 dt = [3t^4/4]_0^1 = 3/4$. This is the first segment of the calculation for $\alpha$.

Is there a shortcut for α?

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Not a direct one, but recognizing it as the sum of areas of “curved triangles” helps in visualizing the process. Each segment adds a specific value to the total.

Why did the factor of 2 appear in Step 3?

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Using the property $\int_0^\pi f(x) dx = 2 \int_0^{\pi/2} f(x) dx$ for symmetric functions (where $f(\pi-x) = f(x)$), which simplifies the upper limit.

What if θ was in degrees?

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Calculus operations like differentiation and integration are strictly performed in radians. If given in degrees, you must convert using $\pi/180$ first.

Why did t⁶ + 1 factorize that way?

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It follows the sum of cubes formula: $a^3 + b^3 = (a+b)(a^2 – ab + b^2)$ where $a = t^2$ and $b = 1$.

Can we solve the trig integral using sin 2θ?

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Yes, substituting $1 – 3\sin^2\theta\cos^2\theta = 1 – \frac{3}{4}\sin^2 2\theta$ is another valid pathway, though $\tan \theta$ is often more direct for high-degree terms.

Is α always an integer in such problems?

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In JEE Main questions, the limits are usually designed so that intermediate values like $\alpha$ are clean integers, making the second part of the question feasible.

What is the range of x¹/³ – [x¹/³]?

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Like all fractional part functions, its range is $[0, 1)$. It never reaches 1.

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