Animated Solution for Mathematics - Definite Integration: The integral ${\int }_0^{\pi }\frac{(x+3)\sin x}{1+3{\cos }^{2}x}dx$ is equal to :

Imagine you are standing before a complex integral, staring at the expression:

At first glance, it looks intimidating. You have a linear term $(x+3)$ multiplying a trigonometric function in the numerator, and a quadratic trigonometric term in the denominator.

If you try to jump straight into integration by parts, you will likely find yourself trapped in a labyrinth of increasingly complex expressions. But in the world of JEE Advanced, there is almost always a hidden symmetry waiting to be exploited.

The $x$ in the numerator is our 'troublemaker.' It is the obstacle preventing a simple substitution. Our goal is to eliminate it, and to do that, we need a secret weapon: King's Property.

King's Property is the most elegant tool in your calculus arsenal. It states that for any definite integral, ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$.

In our case, the limits are $0$ and $\pi$. So, if we replace $x$ with $(\pi -x)$, the value of the integral remains unchanged. Let's see what happens when we apply this transformation:

Now, let's invoke our trigonometric identities. We know that $\sin (\pi -x)=\sin x$ and $\cos (\pi -x)=-\cos x$.

Because the cosine term is squared, the negative sign vanishes, leaving us with ${\cos }^2(\pi -x)={\cos }^{2}x$. Our integral transforms into:

This is where the magic happens. We now have two expressions for $I$. Let's call the original integral equation (1) and our new transformed integral equation (2).

If we add them together, we get $2I$ on the left side. On the right side, because the denominators are identical, we can simply add the numerators:

Look closely at the numerator: $(x+3+\pi -x+3)$. The $x$ and $-x$ cancel out perfectly! We are left with $(\pi +6)$.

The 'troublemaker' is gone, and we are left with a much simpler integral:

Substitution

We can pull the constant $(\pi +6)$ outside. Furthermore, since the function $f(x)=\frac{\sin x}{1+3{\cos }^{2}x}$ satisfies $f(\pi -x)=f(x)$, we can use the symmetry property ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$ to halve the upper limit and multiply by two.

This cancels the $2$ on the left side, giving us:

Now, we use the substitution method. Let $t=\sqrt{3}\cos x$. Then $dt=-\sqrt{3}\sin xdx$, which means $\sin xdx=-\frac{dt}{\sqrt{3}}$.

Changing the limits: when $x=0$, $t=\sqrt{3}$; when $x=\pi /2$, $t=0$. The integral becomes:

This is a standard integral! The integral of $\frac{1}{1+t^2}$ is ${\tan }^{-1}t$. Evaluating this from $0$ to $\sqrt{3}$ gives us ${\tan }^{-1}(\sqrt{3})-{\tan }^{-1}(0)=\frac{\pi }{3}-0=\frac{\pi }{3}$.

Multiplying by our constant, we arrive at the final answer:

And there you have it—a complex problem solved through the elegance of symmetry and substitution. Never fear the $x$ in the numerator; it is often just an invitation to use the King's Property.