Animated Solution for Mathematics - Definite Integration: The value of the definite integral ${\int }_{-\pi /4}^{\pi /4}\frac{dx}{(1+e^{x\cos x})({\sin }^{4}x+{\cos }^{4}x)}$ is equal to:

Imagine you are standing before a massive, complex integral:

At first glance, it looks like a nightmare. You have an exponential function $e^{x\cos x}$ tangled with trigonometric functions in the denominator.

It is easy to feel overwhelmed, but in the world of JEE Advanced, the most intimidating problems often hide the most elegant solutions. The key is to stop looking at the complexity and start looking for the structure.

Whenever you see symmetric limits like $[-\frac{\pi }{4},\frac{\pi }{4}]$, your mind should immediately jump to the King's Property:

Here, $a+b=0$, so the property tells us we can replace $x$ with $-x$. Let's see what happens.

The trigonometric part $({\sin }^{4}x+{\cos }^{4}x)$ is even, meaning it remains unchanged when $x$ becomes $-x$. The exponential term, however, transforms from $e^{x\cos x}$ to $e^{-x\cos x}$. This is the breakthrough we need.

By rewriting $e^{-x\cos x}$ as $\frac{1}{e^{x\cos x}}$, we can manipulate the integral. When we add our original integral $I$ to the transformed integral, the numerators combine to $(1+e^{x\cos x})$, which perfectly cancels the denominator!

Suddenly, the "scary" exponential term has vanished, leaving us with a much friendlier trigonometric integral:

Since the integrand is an even function, we can simplify this to:

Now we are in familiar territory. To solve ${\int }_0^{\pi /4}\frac{dx}{{\sin }^{4}x+{\cos }^{4}x}$, we use the standard trick: divide the numerator and denominator by ${\cos }^{4}x$.

This gives us:

We know that ${\sec }^{4}x={\sec }^{2}x\cdot {\sec }^{2}x=(1+{\tan }^{2}x){\sec }^{2}x$. By substituting $t=\tan x$, we get $dt={\sec }^{2}xdx$, and our integral becomes:

We are almost there. To solve ${\int }_0^1\frac{1+t^2}{1+t^4}dt$, we divide the numerator and denominator by $t^2$, resulting in:

We recognize that the numerator is the derivative of $t-\frac{1}{t}$. We rewrite the denominator as $(t-\frac{1}{t}{)}^2+2$.

Now, let $z=t-\frac{1}{t}$. As $t$ goes from $0$ to $1$, $z$ goes from $-\infty$ to $0$. The integral becomes:

This is a standard form: $\int \frac{dz}{z^2+a^2}=\frac{1}{a}{\tan }^{-1}(\frac{z}{a})$. Applying our limits, we get:

Evaluating this, we find:

We have conquered the beast! The final answer is $\frac{\pi }{2\sqrt{2}}$. Remember, no matter how complex the problem, break it down, look for symmetry, and trust your tools.