Animated Solution for Mathematics - Definite Integration: If ${\int }_{-\pi /2}^{\pi /2}\frac{8\sqrt{2}\cos xdx}{(1+e^{\sin x})(1+{\sin }^{4}x)}=\alpha \pi +\beta {\log }_e(3+2\sqrt{2})$, where $\alpha ,\beta$ are integers, then ${\alpha }^2+{\beta }^2$ equals

Conquering the Exponential Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we stand before a problem that looks like a fortress of complexity. We have an integral with an exponential term in the denominator, a trigonometric function, and a quartic power.

It is designed to intimidate. But remember, in the world of JEE Advanced, intimidation is just a mask for elegance. Let us peel back that mask together.

The King's Gambit

We begin with our integral:

Whenever you see an integral with symmetric limits like $[-\pi /2,\pi /2]$, your mind should immediately race to the King's Property: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$.

Here, $a+b=0$, so we replace $x$ with $-x$. Watch what happens to the exponential term. Since $\sin (-x)=-\sin x$, the term $e^{\sin x}$ transforms into $e^{-\sin x}$.

The $\cos x$ remains $\cos x$ because it is an even function, and ${\sin }^{4}x$ remains unchanged. We now have two expressions for $I$. Adding them is the masterstroke.

The term $\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}$ simplifies beautifully to $1$. The exponential monster is slain.

The Even Function Symmetry

After adding the integrals, we are left with:

Since our integrand is an even function, we can simplify the limits from $[-\pi /2,\pi /2]$ to $[0,\pi /2]$ by multiplying by $2$. This gives us:

This is a much friendlier landscape. We see $\cos xdx$ in the numerator, which is the derivative of $\sin x$. Let $t=\sin x$. Our integral transforms into an algebraic one:

The Algebraic Twins

Now, we face the classic integral $\int \frac{1}{t^4+1}dt$. The trick here is to divide the numerator and denominator by $t^2$.

We rewrite the numerator as $\frac{1}{2}[(1+1/t^2)-(1-1/t^2)]$. This splits our integral into two parts, $I_1$ and $I_2$, each corresponding to a standard form.

For $I_1$, we use the substitution $u=t-1/t$, and for $I_2$, we use $v=t+1/t$. The denominators complete the square to become $(t-1/t{)}^2+(\sqrt{2}{)}^2$ and $(t+1/t{)}^2-(\sqrt{2}{)}^2$.

This is the heart of the problem—linking the numerator to the derivative of the denominator's base.

The Final Victory

Evaluating $I_1$ gives us $\frac{\pi }{2\sqrt{2}}$, and evaluating $I_2$ involves a logarithmic term: $\frac{1}{2\sqrt{2}}\ln (3-2\sqrt{2})$.

With a bit of rationalization, we find that $\ln (3-2\sqrt{2})=-\ln (3+2\sqrt{2})$. Putting it all together, the constants cancel out with the $4\sqrt{2}$ we pulled out earlier, leaving us with $2\pi +2\ln (3+2\sqrt{2})$.

Comparing this to $\alpha \pi +\beta {\log }_e(3+2\sqrt{2})$, we find $\alpha =2$ and $\beta =2$.

The final result, ${\alpha }^2+{\beta }^2=8$, is not just a number; it is the reward for your persistence. You have navigated the symmetry, the substitution, and the algebraic manipulation. That is the beauty of mathematics—the path is difficult, but the destination is pure harmony.