Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{-\pi /2}^{\pi /2}\frac{dx}{(1+e^x)({\sin }^{6}x+{\cos }^{6}x)}$ is equal to

My dear student, welcome to the arena. Today, we are going to dissect a problem that, at first glance, looks like a nightmare. You see an exponential term $e^x$ sitting in the denominator, coupled with a high-power trigonometric expression ${\sin }^{6}x+{\cos }^{6}x$.

It is natural to feel a moment of hesitation. But in the world of JEE Advanced, intimidation is just a sign that you are about to learn something profound. Let us peel back the layers of this problem together.

Imagine you are standing before a locked door. You have the key, but you have to know how to turn it. In definite integration, when you see symmetric limits like $[-\frac{\pi }{2},\frac{\pi }{2}]$, that key is almost always the King's Property:

Here, $a+b=-\frac{\pi }{2}+\frac{\pi }{2}=0$. So, we replace $x$ with $-x$. Our integral becomes:

Because ${\sin }^6(-x)={\sin }^{6}x$ and ${\cos }^6(-x)={\cos }^{6}x$, the trigonometric part remains unchanged. But look at the exponential part! $e^{-x}$ is just $\frac{1}{e^x}$.

When we simplify the denominator $(1+\frac{1}{e^x})$, we get $\frac{e^x+1}{e^x}$. This $e^x$ jumps to the numerator. Now, we have two versions of $I$.

When we add them, the $1+e^x$ terms cancel out perfectly. It is like watching a complex machine suddenly click into place. We are left with:

Now that the exponential term is gone, we are left with a purely trigonometric integral. Because ${\sin }^{6}x+{\cos }^{6}x$ is an even function, we can simplify the limits from $[-\frac{\pi }{2},\frac{\pi }{2}]$ to $[0,\frac{\pi }{2}]$ by multiplying by $2$.

The $2$ on both sides cancels out, leaving us with:

To solve this, we use the standard technique of dividing the numerator and denominator by ${\cos }^{6}x$. This transforms the expression into:

We can rewrite ${\sec }^{6}x$ as $({\sec }^{2}x{)}^2\cdot {\sec }^{2}x$, which is $(1+{\tan }^{2}x{)}^2\cdot {\sec }^{2}x$. This is the perfect setup for a substitution.

Let $t=\tan x$. Then $dt={\sec }^{2}xdx$. As $x$ goes from $0$ to $\frac{\pi }{2}$, $t$ goes from $0$ to $\infty$. Our integral becomes:

Using the sum of cubes formula, $t^6+1=(t^2{)}^3+1^3=(t^2+1)(t^4-t^2+1)$. One factor of $(t^2+1)$ cancels out, leaving us with:

This is a classic biquadratic form. We divide the numerator and denominator by $t^2$:

We are almost there. Let $u=t-\frac{1}{t}$. Then $du=(1+\frac{1}{t^2})dt$.

Also, $u^2=t^2+\frac{1}{t^2}-2$, so $t^2+\frac{1}{t^2}=u^2+2$. The denominator becomes $u^2+2-1=u^2+1$. The limits change from $0$ to $\infty$ for $t$ to $-\infty$ to $\infty$ for $u$.

This is the standard integral for ${\tan }^{-1}u$. Evaluating this from $-\infty$ to $\infty$ gives us:

And there it is. The complexity dissolves into a simple, elegant $\pi$. Remember, in JEE Advanced, the path is rarely a straight line, but if you trust the properties and keep your algebraic steps clean, the solution will always reveal itself.