Animated Solution for Mathematics - Definite Integration: If ${\int }_0^{\pi }\frac{5^{\cos x}(1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x)}{1+5^{\cos x}}dx=\frac{k\pi }{16}$, then $k$ is equal to ______.

The given integral is:

This expression appears daunting due to the exponential term $5^{\cos x}$. However, in JEE Advanced mathematics, such structures are specifically designed to be solved using the King's Property.

The King's Property states that ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$. Applying this to our integral with $a=\pi$, we replace $x$ with $\pi -x$.

Note that $\cos (\pi -x)=-\cos x$. Consequently, the exponential term transforms as follows:

Substituting this into the integral, we get:

Let $P(x)=1+\cos x\cos 3x+{\cos }^{2}x+{\cos }^{3}x\cos 3x$. We must evaluate $P(\pi -x)$:

Since $\cos (\pi -x)=-\cos x$ and $\cos (3\pi -3x)=-\cos 3x$, the product $\cos (\pi -x)\cos (3\pi -3x)$ becomes $(-\cos x)(-\cos 3x)=\cos x\cos 3x$. Thus, $P(\pi -x)=P(x)$.

Adding the original integral $I$ to the transformed integral, the denominators match, and the numerators sum to $P(x)(5^{\cos x}+1)$:

We now evaluate the integral of the polynomial:

1. The integral of $1$ over $[0,\pi ]$ is $\pi$. 2. The integral of ${\cos }^{2}x$ is ${\int }_0^{\pi }\frac{1+\cos 2x}{2}dx=\frac{\pi }{2}$. 3. The integral of $\cos x\cos 3x$ is $0$ due to the orthogonality of trigonometric functions over the interval $[0,\pi ]$.

For the final term, we use ${\cos }^{3}x=\frac{\cos 3x+3\cos x}{4}$:

Summing these results:

Therefore, the value of the integral is:

The final result is $13\pi /16$, implying $k=13$.