Animated Solution for Mathematics - Indefinite Integration: If $I(x)=\int e^{{\sin }^{2}x}(\cos x\sin 2x-\sin x)dx$ and $I(0)=1$, then $I\left(\frac{\pi }{3}\right)$ is equal to

Imagine you are standing before a complex, intimidating integral:

It looks like a tangled mess of trigonometric functions and exponential growth. In the world of JEE Advanced, intimidation is just a sign that you haven't yet seen the underlying symmetry. Let's embark on a journey to uncover it.

The first thing that should catch your eye is the exponent: ${\sin }^{2}x$. Whenever you see a function in the exponent of $e$, your mathematical intuition should immediately suggest a substitution.

Let us set $t={\sin }^{2}x$. Now, we need to find $dt$. Differentiating $t$ with respect to $x$, we get:

Using the double angle identity, this is simply $\sin 2x$. Thus, $dt=\sin 2xdx$. This is our golden ticket.

Now, look at the integrand: $(\cos x\sin 2x-\sin x)$. We have a $\sin 2x$ term, but we also have a pesky $-\sin x$.

To make this work with our $dt=\sin 2xdx$, we need to express everything in terms of $\sin 2x$. We can rewrite $\sin x$ as:

Now, our integrand becomes:

This is the breakthrough! We have successfully factored out the $\sin 2x$ that we need for our $dt$.

With $t={\sin }^{2}x$, we know that ${\cos }^{2}x=1-t$, so $\cos x=\sqrt{1-t}$. Substituting this into our integral, we get:

Does this look familiar? It is the classic form $\int e^t(f(t)+f^{\prime }(t))dt$.

If we let $f(t)=\sqrt{1-t}$, then its derivative $f^{\prime }(t)$ is indeed:

The integral of this form is simply $e^{t}f(t)+C$.

Applying this, we get $I(x)=e^t\sqrt{1-t}+C$. Substituting back $t={\sin }^{2}x$ and $\sqrt{1-t}=\cos x$, we arrive at:

Using the initial condition $I(0)=1$, we find $C=0$. Finally, evaluating at $x=\frac{\pi }{3}$, we have ${\sin }^2(\frac{\pi }{3})=\frac{3}{4}$ and $\cos (\frac{\pi }{3})=\frac{1}{2}$.

The final result is:

You see? What started as a terrifying expression was just a beautiful, hidden structure waiting for you to reveal it. Keep practicing, and you will start seeing these patterns everywhere!