Animated Solution for Mathematics - Definite Integration: Let $f(x)=x+\frac{a}{{\pi }^2-4}\sin x+\frac{b}{{\pi }^2-4}\cos x,x\in \mathbb{R}$ be a function which satisfies $f(x)=x+{\int }_0^{\pi /2}\sin (x+y)f(y)dy$. Then $(a+b)$ is equal to

Welcome, future engineer! Today, we are tackling a problem that often intimidates students: the integral equation.

When you see the equation: it is natural to feel a bit overwhelmed. It looks like a recursive loop, but it is actually a puzzle waiting to be disassembled.

The core of this problem lies in the kernel $\sin (x+y)$. In the world of integral equations, whenever you see a compound angle inside an integral, your first instinct should be to break it apart using the identity $\sin (x+y)=\sin x\cos y+\cos x\sin y$.

We can rewrite our equation as:

Suddenly, the variables $x$ and $y$ are no longer tangled. We can distribute $f(y)$ and pull the terms involving $x$ outside the integral, because they are constant with respect to $y$:

Now, let's perform the magic trick. The integrals ${\int }_0^{\pi /2}\cos yf(y)dy$ and ${\int }_0^{\pi /2}\sin yf(y)dy$ are just constants. Let's define them as $A$ and $B$, respectively.

Our equation becomes:

This is a massive breakthrough! We have reduced the entire integral equation to finding two constants, $A$ and $B$.

By comparing this to the given form $f(x)=x+\frac{a}{{\pi }^2-4}\sin x+\frac{b}{{\pi }^2-4}\cos x$, we identify that $A=\frac{a}{{\pi }^2-4}$ and $B=\frac{b}{{\pi }^2-4}$. Our goal is to find $a+b$, which is simply $({\pi }^2-4)(A+B)$.

Let $K=A+B$. If we can find $K$, we have solved the problem.

To find $K$, we evaluate the sum of the integrals:

Substituting our expression for $f(y)$, we get:

We split this into two parts: the integral involving $y$ and the integral involving $A$ and $B$. The integral ${\int }_0^{\pi /2}y(\sin y+\cos y)dy$ evaluates to $\frac{\pi }{2}$ after integration by parts.

The second part, involving $K$, simplifies to $K\cdot I_1$, where:

We are left with the equation:

Rearranging the terms to solve for $K$:

Finally, we calculate $a+b=({\pi }^2-4)K$. Substituting $K=-1$:

You have successfully navigated the integral, the algebra, and the calculus. Take a moment to appreciate the symmetry and the structure—this is the heart of JEE mathematics!