Animated Solution for Mathematics - Definite Integration: If ${\int }_0^{\pi }({\sin }^{3}x)e^{-{\sin }^{2}x}dx=\alpha -\frac{\beta }{e}{\int }_0^1\sqrt{t}{e}^{t}dt$, then $\alpha +\beta$ is equal to

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving an integral; we are conducting a symphony.

When you first look at the expression $I={\int }_0^{\pi }({\sin }^{3}x)e^{-{\sin }^{2}x}dx$, it is natural to feel a momentary shiver. It looks like a chaotic mess of trigonometric and exponential functions.

But in the world of advanced calculus, chaos is often just order in disguise. Let us peel back the layers together.

Whenever you see an integral with limits from $0$ to $\pi$, your internal alarm bells should ring. This is the domain of symmetry.

We know that $\sin (\pi -x)=\sin x$. Because our integrand $f(x)={\sin }^{3}x{e}^{-{\sin }^{2}x}$ satisfies $f(\pi -x)=f(x)$, the function is perfectly symmetric about $x=\frac{\pi }{2}$.

This is a gift! It allows us to rewrite the integral as:

By halving the interval and doubling the result, we have already simplified our mental burden. We are now working in the first quadrant, where everything is positive and well-behaved.

Now, let us look at the integrand: ${\sin }^{3}x{e}^{-{\sin }^{2}x}$. We need to make this look like something we can handle. Let us use the identity ${\sin }^{2}x=1-{\cos }^{2}x$.

First, ${\sin }^{3}x$ becomes $\sin x(1-{\cos }^{2}x)$. Second, the exponential term $e^{-{\sin }^{2}x}$ transforms into $e^{-(1-{\cos }^{2}x)}=e^{-1}\cdot e^{{\cos }^{2}x}$.

Substituting these back, we pull the constant $e^{-1}$ out, and our integral becomes:

Do you see it? The $\sin x$ is waiting patiently to be the derivative of $\cos x$. This is the moment where we apply the substitution $u=\cos x$.

Then $du=-\sin xdx$. As $x$ goes from $0$ to $\frac{\pi }{2}$, $u$ goes from $1$ to $0$. The negative sign from $du$ flips the limits back to $0$ to $1$. We are left with:

We have arrived at a point where we must split the integral into two parts: ${\int }_0^1{e}^{u^2}du$ and ${\int }_0^1{u}^2{e}^{u^2}du$. The first part is a non-elementary integral—we cannot solve it in terms of basic functions.

But that is okay! The problem asks us to relate our answer to another integral. We focus our energy on the second part, $J={\int }_0^1{u}^2{e}^{u^2}du$.

To solve $J$, we use Integration by Parts. We rewrite $u^2$ as $\frac{u}{2}\cdot (2u)$. Why? Because $2u$ is the derivative of $u^2$, which makes the exponential term $e^{u^2}$ easy to integrate.

Let $f=\frac{u}{2}$ and $g^{\prime }=2u{e}^{u^2}$. Then $f^{\prime }=\frac{1}{2}$ and $g=e^{u^2}$. Applying the formula $\int f{g}^{\prime }=fg-\int f^{\prime }g$, we get:

Now, we substitute $J$ back into our expression for $I$. After some algebraic cleanup, we find:

Simplifying this expression yields:

But the question gives us a target: $\alpha -\frac{\beta }{e}{\int }_0^1\sqrt{t}{e}^{t}dt$. We need to connect ${\int }_0^1{e}^{u^2}du$ to ${\int }_0^1\sqrt{t}{e}^{t}dt$.

Let $K={\int }_0^1\sqrt{t}{e}^{t}dt$. Using the substitution $t=u^2$, we have $dt=2udu$, which transforms $K$ into ${\int }_0^{1}u{e}^{u^2}(2u)du=2{\int }_0^1{u}^2{e}^{u^2}du$.

Alternatively, using Integration by Parts on $K$ (with $f=\sqrt{t}$ and $g^{\prime }=e^t$), we find that ${\int }_0^1{e}^{u^2}du$ relates directly to the target form. Following the substitution logic, we arrive at:

Comparing this to $\alpha -\frac{\beta }{e}K$, we see clearly that $\alpha =2$ and $\beta =3$. The final result is: