Evaluate the following integral: $I=\int{\sin 4x\cdot {{e}^{{{\tan }^{2}}x}}dx}$.

Hint: We will simplify the function that we have to integrate. We will do this by using trigonometric identities such as $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ and $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$ . Then to integrate, we will use the method of substitution. We will come across a step resembling the form ${{e}^{x}}\left( f(x)+{f}'(x) \right)$ and we know that $\int{{{e}^{x}}\left( f(x)+{f}'(x) \right)dx={{e}^{x}}f(x)+c}$ where $c$ is the integration constant. Then we will substitute the value of the substituted variable to obtain the final answer.

Complete step-by-step answer:
We have to integrate the following: $I=\int{\sin 4x\cdot {{e}^{{{\tan }^{2}}x}}dx}$.
Now, we will write $\sin (4x)=\sin 2\cdot (2x)=2\sin 2x\cos 2x$. So we get the following expression,
$I=\int{2\sin 2x\cos 2x\cdot {{e}^{{{\tan }^{2}}x}}dx}$.
We will substitute the values $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ and $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$ in the above equation. We get the following,
 $I=\int{2\cdot \dfrac{2\tan x}{1+{{\tan }^{2}}x}\cdot \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}\cdot {{e}^{{{\tan }^{2}}x}}dx}$.
Simplifying this equation, we get,
$I=\int{2\cdot \dfrac{1-{{\tan }^{2}}x}{{{\left( 1+{{\tan }^{2}}x \right)}^{2}}}\cdot {{e}^{{{\tan }^{2}}x}}\cdot 2\tan xdx}$
Now, we will substitute ${{\tan }^{2}}x=t$, therefore we have $2\tan x{{\sec }^{2}}xdx=dt$, that is $2\tan xdx=\dfrac{dt}{{{\sec }^{2}}x}$ . Substituting these values in the above equation, we get
$I=\int{2\cdot \dfrac{1-t}{{{(1+t)}^{2}}}\cdot {{e}^{t}}\dfrac{dt}{{{\sec }^{2}}x}}$ .
But we know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x=1+t$. Therefore, we have
$I=\int{2{{e}^{t}}\dfrac{1-t}{{{(1+t)}^{3}}}dt}$ .
Now we will add and subtract 1 in the numerator, since it will not change the value of the function, in the following manner,
\[\begin{align}
  & I=\int{2{{e}^{t}}\dfrac{1+1-1-t}{{{(1+t)}^{3}}}dt} \\
 & =\int{2{{e}^{t}}\dfrac{2-(1+t)}{{{(1+t)}^{3}}}dt} \\
 & =\int{2{{e}^{t}}\left[ \dfrac{2}{{{(1+t)}^{3}}}-\dfrac{1+t}{{{(1+t)}^{3}}} \right]dt} \\
 & =\int{2{{e}^{t}}\left[ \dfrac{2}{{{(1+t)}^{3}}}-\dfrac{1}{{{(1+t)}^{2}}} \right]dt}
\end{align}\]
Here, note that $f(t)=-\dfrac{1}{{{(1+t)}^{2}}}$ and ${f}'(t)=\dfrac{2}{{{(1+t)}^{3}}}$ . We know that $\int{{{e}^{x}}\left( f(x)+{f}'(x) \right)dx={{e}^{x}}f(x)+c}$. Using this result, we get
$I=2\cdot {{e}^{t}}\cdot -\dfrac{1}{{{(1+t)}^{2}}}+c$ where $c$ is the integration constant.
Now, we have to re substitute ${{\tan }^{2}}x=t$. We get the following expression,
$I=-\dfrac{2{{e}^{{{\tan }^{2}}x}}}{{{(1+{{\tan }^{2}}x)}^{2}}}+c$.

Note: As the function to be integrated is trigonometric, we should aim to simplify the function using trigonometric identities. The method of integration by substitution looks like an intuitive choice after looking at the multiple ${{\tan }^{2}}x$ terms in the integrand. Use of trigonometric identities needs to be done carefully because it is possible to get caught in loops.