Animated Solution for Mathematics - Definite Integration: The function $f(x)$, that satisfies the condition $f(x)=x+{\int }_0^{\pi /2}\sin x\cdot \cos yf(y)dy$, is:

Welcome, my dear student. Today, we are going to peel back the layers of a problem that often intimidates students at first glance: the Fredholm integral equation. It looks like a loop, doesn't it?

We have $f(x)$ defined by an integral that contains $f(y)$. It feels like a snake eating its own tail. But I want you to take a deep breath. In the world of JEE Advanced, these problems are not designed to trap you; they are designed to test your ability to see the hidden structure beneath the complexity.

The Constant Insight

Let us look at the equation:

The first thing we must do is observe the variable of integration. We are integrating with respect to $y$. This is the key to the entire kingdom.

Because the integral is strictly about $y$, the term $\sin x$ is completely oblivious to the integration process. To the integral, $\sin x$ is just a constant, like the number $5$ or $\pi$.

So, let us pull it out! We rewrite the equation as:

Suddenly, the equation looks much less intimidating, doesn't it?

The Power of Substitution

Now, look at that integral: ${\int }_0^{\pi /2}\cos yf(y)dy$. Since the limits are $0$ and $\pi /2$, this integral is not a function of $x$; it is a fixed, numerical value.

Let us call this value $K$. By defining $K={\int }_0^{\pi /2}\cos yf(y)dy$, we have transformed our complex integral equation into a simple algebraic one:

This is the 'soul' of our function. We now know that $f(x)$ must be of the form $x$ plus some constant times $\sin x$. Our only mission now is to find that constant $K$.

The Recursive Loop

How do we find $K$? We use the definition we just created! We know $K={\int }_0^{\pi /2}\cos yf(y)dy$.

Since we now know that $f(y)=y+K\sin y$, we can substitute this directly into our integral. This gives us:

By distributing the $\cos y$, we get:

Using the linearity of integrals, we split this into two parts:

The Calculus Journey

Let us solve these two integrals one by one. For the first part, $I_1={\int }_0^{\pi /2}y\cos ydy$, we use integration by parts.

Let $u=y$ and $dv=\cos ydy$. Then $du=dy$ and $v=\sin y$. The formula $\int udv=uv-\int vdu$ gives us:

Evaluating this, we get $(\frac{\pi }{2}\cdot 1-0)-[-\cos y{]}_0^{\pi /2}=\frac{\pi }{2}-(0-(-1))=\frac{\pi }{2}-1$. Beautiful, isn't it?

Now for the second part, $I_2=K{\int }_0^{\pi /2}\sin y\cos ydy$. We know that $\sin y\cos y=\frac{1}{2}\sin (2y)$.

Integrating this, we get:

The Final Victory

We are almost there. Putting it all together, we have:

Subtracting $\frac{K}{2}$ from both sides, we get $\frac{K}{2}=\frac{\pi }{2}-1$. Multiplying by $2$, we find $K=\pi -2$.

We have conquered the constant! Now, we simply substitute $K$ back into our structural equation:

You have successfully navigated the integral equation. Take a moment to appreciate the elegance of the result. You didn't just solve a problem; you decoded a mathematical structure. Keep this confidence with you for the next one!