If y(x) = ∫(π²/16 to x²) (cos x cos √θ) / (1 + sin² √θ) dθ, then find dy/dx at x = π.

Q: Wait, why can't I just integrate the function first?

That is a very natural instinct! However, in JEE Advanced problems, the integral is often designed to be non-elementary or extremely tedious to solve directly. The problem is testing your ability to use the properties of derivatives and the Leibniz Rule, not your ability to perform brute-force integration. Always look for a shortcut before diving into the integral.

Q: Why did we pull $\cos x$ out of the integral?

This is the most crucial observation! The integral is defined with respect to $\theta$. If a term inside the integral does not contain $\theta$, it is treated as a constant relative to the integration process. By pulling $\cos x$ out, we transform a complex expression into a simple product of two functions, which is much easier to handle.

Q: What happens to the lower limit in the Leibniz rule?

The Leibniz Rule states that the derivative of $\int_{a}^{g(x)} f(\theta) d\theta$ is $f(g(x)) \cdot g'(x) - f(a) \cdot a'$. Since our lower limit is the constant $\frac{\pi^2}{16}$, its derivative is zero. Therefore, the lower limit contributes nothing to the final derivative. You can safely ignore it!

Question

If $y(x)=\int_{\pi^{2} / 16}^{x^{2}} \frac{\cos x \cos \sqrt{	heta}}{1+\sin ^{2} \sqrt{	heta}} d 	heta$, then find $\frac{d y}{d x}$ at $x=\pi$.

Sigma Percentile · Accepted Answer

### The Arena of Calculus

Welcome, student. Today, we stand before a problem that looks like a daunting fortress. At first glance, you see an integral, a trigonometric function, and a variable $x$ lurking in both the limits and the integrand.

It is designed to intimidate. But I want you to take a deep breath. In the world of JEE Advanced, we do not fight monsters with brute force; we dismantle them with elegance.

### The Art of Observation

Our function is defined as:

$$y(x) = \int_{\frac{\pi^2}{16}}^{x^2} \frac{\cos x \cos \sqrt{	heta}}{1+\sin^2 \sqrt{	heta}} d	heta$$

The first thing to notice is the variable of integration: $	heta$. Everything else is just scenery. Look at that $\cos x$ in the numerator; it does not depend on $	heta$.

It is completely independent. This is our 'Aha!' moment. We can treat $\cos x$ as a constant relative to $	heta$ and pull it outside the integral sign.

Suddenly, our function becomes:

$$y(x) = \cos x \cdot \int_{\frac{\pi^2}{16}}^{x^2} \frac{\cos \sqrt{	heta}}{1+\sin^2 \sqrt{	heta}} d	heta$$

We have turned a complex integral into a simple product of two functions, $u(x) = \cos x$ and $v(x) = \int_{\frac{\pi^2}{16}}^{x^2} \frac{\cos \sqrt{	heta}}{1+\sin^2 \sqrt{	heta}} d	heta$.

### The Toolbox: Product Rule and Leibniz

Now that we have a product, the path forward is clear. We need the derivative $\frac{dy}{dx}$. We reach for our trusty Product Rule:

$$\frac{d}{dx}[u \cdot v] = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}$$

The first part, $\frac{du}{dx}$, is easy: the derivative of $\cos x$ is $-\sin x$. The second part, $\frac{dv}{dx}$, requires the **Newton-Leibniz Rule**.

This rule is the secret weapon for differentiating integrals with variable limits. It states:

$$\frac{d}{dx} \int_{a}^{g(x)} f(	heta) d	heta = f(g(x)) \cdot g'(x)$$

We substitute our upper limit $g(x) = x^2$ into the integrand, replacing $	heta$ with $x^2$, and then multiply by the derivative of the limit, which is $2x$.

### The Grand Finale

Let us assemble our pieces. The derivative is:

$$\frac{dy}{dx} = -\sin x \int_{\frac{\pi^2}{16}}^{x^2} \frac{\cos \sqrt{	heta}}{1+\sin^2 \sqrt{	heta}} d	heta + \cos x \left( \frac{\cos x}{1+\sin^2 x} \cdot 2x ight)$$

Now, we evaluate this at $x = \pi$. Here is the beauty of the problem: we know that $\sin \pi = 0$. The entire first term, which contains that 'scary' integral, vanishes into thin air!

We are left with only the second term:

$$\cos \pi \left( \frac{\cos \pi}{1+\sin^2 \pi} \cdot 2\pi ight)$$

Since $\cos \pi = -1$ and $\sin \pi = 0$, this simplifies to $(-1) \cdot \left( \frac{-1}{1+0} \cdot 2\pi ight)$.

The final result is **$2\pi$**. We have conquered the fortress. Remember, in mathematics, the most complex problems often collapse under the weight of a single, elegant observation.

Animated Solution for Mathematics - Definite Integration: If $y (x) = \int_{π^{2} /16}^{x^{2}} \frac{c o s x c o s θ}{1 + s i n ^{2} θ} d θ$ , then find $\frac{d y}{d x}$ at $x = π$ .

Visualized Solution

Analyzing the Function $y (x)$

Pulling out $cos x$

Rewriting $y (x)$

Applying the Product Rule

Differentiating $u (x)$

Differentiating the Integral

Substituting the Upper Limit

Multiplying by $g^{'} (x)$

Combining the Terms

Substituting $x = π$

Evaluating the First Term

Calculating the Second Term

Final Simplification

The Sigma Insight: Newton-Leibniz & Reduction Formulas

The Arena of Calculus

The Art of Observation

The Toolbox

The Grand Finale

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(x)=∫π2/16x2​1+sin2θ​cosxcosθ​​dθ, then find dxdy​ at x=π.

Visualized Solution

Analyzing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(x)

Pulling out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosx

Rewriting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(x)