lim(x→π/2) [∫[x³ to (π/2)³] (sin(2t^(1/3)) + cos(t^(1/3))) dt] / (x - π/2)² is equal to
(1) 5π²/9
(2

Q: Wait, why do we get $\frac{0}{0}$ again after applying L'Hopital's Rule?

That is a classic JEE trap! L'Hopital's Rule only resolves one layer of the indeterminate form. In this problem, the derivative of the numerator still contains a $\cos x$ term, and the denominator still contains $(x - \frac{\pi}{2})$. Since $\cos(\frac{\pi}{2}) = 0$ and $(\frac{\pi}{2} - \frac{\pi}{2}) = 0$, the form persists. Instead of applying L'Hopital again—which would lead to a messy product rule—we use trigonometric identities to isolate the standard limit $\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{x - \frac{\pi}{2}} = -1$.

Q: Can I just evaluate the integral first and then take the limit?

In theory, yes, but in practice, it is a nightmare. The function $\sin(2t^{1/3}) + \cos(t^{1/3})$ does not have an elementary antiderivative that is easy to work with. The beauty of the Newton-Leibniz formula is that it allows us to bypass the integration entirely. Always look for ways to differentiate the integral rather than solving it when you see it inside a limit!

Q: Why did the upper limit of the integral disappear?

Great observation! The Newton-Leibniz formula requires us to multiply the function by the derivative of the limit. Our upper limit was $(\frac{\pi}{2})^3$, which is a constant. The derivative of any constant is $0$. Therefore, the entire term associated with the upper limit vanishes, leaving us only with the lower limit's contribution.

Question

$\lim_{x 	o \frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} (\sin(2t^{1/3}) + \cos(t^{1/3})) dt}{(x - \frac{\pi}{2})^2}$ is equal to

Options:

(A) $\frac{5\pi^2}{9}$

(B) $\frac{9\pi^2}{8}$

(C) $\frac{11\pi^2}{10}$

(D) $\frac{3\pi^2}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex limit, staring at an integral trapped in the numerator. It looks intimidating, but in the world of JEE Advanced, intimidation is often just a mask for elegance.

We are tasked with evaluating the following limit:

$$L = \lim_{x 	o \frac{\pi}{2}} \frac{\int_{x^3}^{(\frac{\pi}{2})^3} (\sin(2t^{\frac{1}{3}}) + \cos(t^{\frac{1}{3}})) dt}{(x - \frac{\pi}{2})^2}$$

### The First Encounter

Before we reach for any heavy machinery, we must respect the ritual of limits: direct substitution. As $x 	o \frac{\pi}{2}$, the lower limit of our integral, $x^3$, approaches $(\frac{\pi}{2})^3$.

When the upper and lower limits of an integral are identical, the area under the curve is zero. Simultaneously, the denominator $(x - \frac{\pi}{2})^2$ also becomes zero.

We have confirmed the $\frac{0}{0}$ indeterminate form. This is our green light to proceed.

### The Newton-Leibniz Weapon

We need to differentiate the numerator, which is an integral. This is where the Newton-Leibniz formula becomes our best friend:

$$\frac{d}{dx} \int_{h(x)}^{g(x)} f(t) dt = f(g(x))g'(x) - f(h(x))h'(x)$$

Applying this to our numerator, we note that the upper limit is a constant, so its derivative is $0$. The lower limit is $h(x) = x^3$, so its derivative is $3x^2$.

Substituting $t = x^3$ into our integrand, we get $\sin(2(x^3)^{\frac{1}{3}}) + \cos((x^3)^{\frac{1}{3}})$, which simplifies beautifully to $\sin(2x) + \cos x$. Thus, the derivative of our numerator is:

$$-3x^2(\sin(2x) + \cos x)$$

### The Second Hurdle

Now, we differentiate the denominator. Using the power rule, the derivative of $(x - \frac{\pi}{2})^2$ is simply $2(x - \frac{\pi}{2})$.

Assembling our new limit, we have:

$$\lim_{x 	o \frac{\pi}{2}} \frac{-3x^2(\sin(2x) + \cos x)}{2(x - \frac{\pi}{2})}$$

If we check the form again, we still have $\frac{0}{0}$. Rather than applying L'Hopital's Rule again, which would involve a messy product rule, let us use our mathematical intuition.

We know that $\sin(2x) = 2\sin x \cos x$. Substituting this, our numerator becomes $-3x^2(2\sin x \cos x + \cos x)$. Factoring out $\cos x$, we get:

$$-3x^2 \cos x (2\sin x + 1)$$

### The Final Victory

Now, look at the expression:

$$\lim_{x 	o \frac{\pi}{2}} \frac{-3x^2 \cos x (2\sin x + 1)}{2(x - \frac{\pi}{2})}$$

We can isolate the standard limit $\lim_{x 	o \frac{\pi}{2}} \frac{\cos x}{x - \frac{\pi}{2}}$. By substituting $\cos x = \sin(\frac{\pi}{2} - x)$, this part evaluates to $-1$.

Now, we simply plug in $x = \frac{\pi}{2}$ for the remaining terms:

$$\frac{-3(\frac{\pi}{2})^2 \cdot (-1) \cdot (2\sin(\frac{\pi}{2}) + 1)}{2}$$

Since $\sin(\frac{\pi}{2}) = 1$, the bracket becomes $2(1) + 1 = 3$. Our calculation becomes:

$$\frac{3 \cdot \frac{\pi^2}{4} \cdot 3}{2} = \mathbf{\frac{9\pi^2}{8}}$$

We have conquered the limit! Remember, the key is not to fear the integral, but to master the tools that allow you to dismantle it.

Animated Solution for Mathematics - Definite Integration: $lim_{x \to \frac{π}{2}} \frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} )) d t}{( x - \frac{π}{2} ) ^{2}}$ is equal to

Select Answer:

Visualized Solution

Identify the Form of the Limit

Checking the Indeterminate Form

Applying L'Hopital's Rule

The Newton-Leibniz Formula

Differentiating the Numerator

Differentiating the Denominator

Assembling the New Limit

Trigonometric Simplification

Handling the Standard Limit

Final Substitution

The Final Answer

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The First Encounter

The Newton-Leibniz Weapon

The Second Hurdle

The Final Victory

Similar Questions

$lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (t^{\frac{1}{3}}) d t)$ is equal to

$lim_{x \to 0} \frac{\int _{0}^{x^{2}} s i n t d t}{x ^{3}}$ is equal to:

$lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

$lim_{x \to 1} \frac{\int _{0}^{(x - 1)^{2}} t c o s ( t ^{2} ) d t}{( x - 1 ) s i n ( x - 1 )}$

$lim_{x \to 0} \frac{\int _{0}^{x} t s i n ( 10 t ) d t}{x}$ is equal to

$lim_{x \to 0} \frac{48}{x ^{4}} \int_{0}^{x} \frac{t ^{3}}{t ^{6} + 1} d t$ is equal to ______.

If $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

Let $f$ be a differentiable function in $(0, \frac{π}{2})$ . If $\int_{c o s x}^{1} t^{2} f (t) d t = sin^{3} x + cos x$ , then $\frac{1}{3} f^{'} (\frac{1}{3})$ is equal to :

If $ϕ (x) = \frac{1}{x} \int_{π /4}^{x} (42 sin t - 3 ϕ^{'} (t)) d t, x > 0$ , then $ϕ^{'} (π /4)$ is equal to :

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→2π​​(x−2π​)2∫x3(π/2)3​(sin(2t1/3)+cos(t1/3))dt​ is equal to