Animated Solution for Mathematics - Definite Integration: The value of ${\int }_0^{2\pi }\frac{x{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$ is equal to:

Imagine you are standing before a complex integral:

At first glance, it looks intimidating. The variable $x$ in the numerator seems to be the source of all our trouble.

In the world of JEE Advanced, whenever you see an $x$ multiplied by a function, there is almost always a hidden symmetry waiting to be exploited. We call this the King's Property.

Our Magic Wand

The King's Property is defined as:

It is the most powerful tool in your arsenal for definite integration. By replacing $x$ with $a-x$, we often find that the troublesome $x$ term cancels out.

In our case, $a=2\pi$. So, we write:

Now, we must be careful. We need to evaluate $\sin (2\pi -x)$ and $\cos (2\pi -x)$.

Since $2\pi -x$ is in the fourth quadrant, $\sin (2\pi -x)=-\sin x$ and $\cos (2\pi -x)=\cos x$. However, because we are raising these to the power of $8$ (an even number), the negative sign vanishes:

The denominator remains unchanged. Our integral now looks like this:

Now, let us add our original integral $I$ to this new form. On the left, we get $2I$.

On the right, since the denominators are identical, we add the numerators:

Look closely at the numerator: $x{\sin }^{8}x+2\pi {\sin }^{8}x-x{\sin }^{8}x$. The $x{\sin }^{8}x$ terms cancel out perfectly!

We are left with:

We can pull the constant $2\pi$ outside:

We are almost there. The function $f(x)=\frac{{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}$ has a period of $\pi /2$.

Using the property ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$, we can reduce the interval $[0,2\pi ]$ to $[0,\pi /2]$ by multiplying by $4$:

Let $J={\int }_0^{\pi /2}\frac{{\sin }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$. Applying the King's Property again to $J$ (with $a=\pi /2$), we get:

Adding these two forms of $J$ gives:

Finally, we calculate the result:

The elegance of this result is a testament to the power of symmetry in mathematics.