Animated Solution for Mathematics - Definite Integration: Let $f(\theta )=\sin \theta +{\int }_{-\pi /2}^{\pi /2}(\sin \theta +t\cos \theta )f(t)dt$. Then the value of ${\int }_0^{\pi /2}f(\theta )d\theta$ is ______.

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a beautiful mathematical structure.

At first glance, this integral equation might seem like a labyrinth:

It looks recursive, but this is a Fredholm integral equation of the second kind, and it is far more approachable than it appears.

The first step is to demystify the integral. Notice that the limits of integration are fixed from $-\pi /2$ to $\pi /2$, meaning the entire integral evaluates to a constant.

We can rewrite the integral as:

Let us define these two integrals as constants:

Suddenly, our equation becomes much friendlier:

This simplifies to:

Now, we need to find $A$ and $B$. Let us look at $A$:

Here, we use the property of odd and even functions. Since $\sin t$ is an odd function, its integral over $[-\pi /2,\pi /2]$ is $0$.

Since $\cos t$ is an even function, its integral is:

Thus, we obtain the relation $A=2B$.

Next, we tackle $B$:

Expanding this, we get:

The second integral involves $t\cos t$, which is an odd function, so it vanishes. The first integral involves $t\sin t$, which is an even function, so we evaluate:

Using integration by parts, $\int t\sin tdt=-t\cos t+\sin t$. Evaluating from $0$ to $\pi /2$ gives $1$. Thus, $B=2(A+1)$.

We now have a simple system:

Substituting $A=2B$ into the second equation gives $B=2(2B)+2$, which leads to $-3B=2$, or $B=-2/3$. Consequently, $A=-4/3$.

With $A=-4/3$ and $B=-2/3$, our function is:

The final task is to integrate this from $0$ to $\pi /2$:

Evaluating the bounds:

We have arrived at the final answer: $-1$. Keep this mindset—always look for the constants hidden in the integrals.