Evaluate: $\int{\sin 4x{{e}^{{{\tan }^{2}}x}}dx}$

Hint: Here, this question is an indefinite integration with trigonometric terms. Here, we can't directly solve such questions, we have to manipulate the terms into a simpler form so that it can be substituted to other simpler form and finally integrate and get an answer. Here, we have ${{\tan }^{2}}x$t as the power of 'e'. Therefore, it is clear that, we have to substitute the ${{\tan }^{2}}x$ as t. But before that, we will manipulate the other terms in such a way that it converts into ${{\tan }^{2}}x$ terms format.

Complete step by step answer:
So, let's get started, we have been given in the question:
Let’s suppose it as \[I=\int{\sin 4x{{e}^{{{\tan }^{2}}x}}}dx\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
We know,
\[\begin{align}
  & \sin 2\theta =2\sin \theta \cos \theta \\
 & \therefore \sin 4x=\sin 2\left( 2x \right)=2\sin \left( 2x \right)\cdot \cos \left( 2x \right) \\
\end{align}\]
Now put in equation (i) we get:
\[I=\int{2\sin 2x\cos 2x\text{ }{{e}^{{{\tan }^{2}}x}}\text{ }}dx\]
Now, we know the formula,
\[\sin 2x=\dfrac{2\tan x}{\left( 1+{{\tan }^{2}}x \right)}\text{ and }\cos 2x=\left( \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)\]
Therefore,
\[\begin{align}
  & I=\int{2\times \left( \dfrac{2\tan x}{1+{{\tan }^{2}}x} \right)}\times \left( \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right){{e}^{{{\tan }^{2}}x}}dx \\
 & I=4\int{\dfrac{\tan x\left( 1-{{\tan }^{2}}x \right)}{{{\left( 1+{{\tan }^{2}}x \right)}^{2}}}}\text{ }{{e}^{{{\tan }^{2}}x}}dx \\
\end{align}\]
Now, put ${{\tan }^{2}}x=t$ by differentiating both sides, we get:
\[\begin{align}
  & 2\tan x{{\sec }^{2}}xdx=dt \\
 & 2\tan xdx=\dfrac{dt}{{{\sec }^{2}}x} \\
 & 2\tan xdx=\dfrac{dt}{\left( 1+{{\tan }^{2}}x \right)}\left( \because {{\sec }^{2}}x=\left( 1+{{\tan }^{2}}x \right) \right) \\
 & \therefore 2\tan xdx=\left( \dfrac{dt}{1+t} \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
Hence,
\[\begin{align}
  & I=4\int{\dfrac{\tan x\left( 1-{{\tan }^{2}}x \right)}{{{\left( 1+{{\tan }^{2}}x \right)}^{2}}}}\text{ }{{e}^{{{\tan }^{2}}x}}dx \\
 & I=2\int{\dfrac{\left( 1-{{\tan }^{2}}x \right)}{{{\left( 1+{{\tan }^{2}}x \right)}^{2}}}}\times {{e}^{{{\tan }^{2}}x}}\left( 2\tan xdx \right) \\
 & I=2\int{\dfrac{\left( 1-t \right)}{{{\left( 1+t \right)}^{2}}}}{{e}^{t}}\left( \dfrac{dt}{1+t} \right)\left( \text{from equation }\left( \text{iii} \right) \right) \\
 & I=2\int{\dfrac{\left( 1-t \right)}{{{\left( 1+t \right)}^{3}}}}{{e}^{t}}dt \\
\end{align}\]
Now, add and subtract 1 in numerator, we get:
\[\begin{align}
  & I=2\int{\dfrac{\left( 1+1-1-t \right)}{{{\left( 1+t \right)}^{3}}}}{{e}^{t}}dt \\
 & I=2\int{\dfrac{2-\left( 1+t \right)}{{{\left( 1+t \right)}^{3}}}}{{e}^{t}}dt \\
 & I=2\int{\left( \dfrac{2}{{{\left( 1+t \right)}^{3}}}-\dfrac{\left( 1+t \right)}{{{\left( 1+t \right)}^{3}}} \right)}{{e}^{t}}dt \\
 & I=2\int{\left( \dfrac{2}{{{\left( 1+t \right)}^{3}}}-\dfrac{1}{{{\left( 1+t \right)}^{2}}} \right)}{{e}^{t}}dt \\
\end{align}\]
Now, we have one formula, i.e. \[\int{{{e}^{x}}}\left( f\left( x \right)+f'\left( x \right) \right)dx={{e}^{x}}f\left( x \right)+c\]
That means, if sum of function and its differentiation written with ${{e}^{x}}$ then output will be product of function and ${{e}^{x}}$
In our case, \[I=2\int{\left( \dfrac{2}{{{\left( 1+t \right)}^{3}}}+\left( \dfrac{-1}{{{\left( 1+t \right)}^{2}}} \right) \right)}{{e}^{t}}dt\]
Here, function will be $\left( \dfrac{-1}{{{\left( 1+t \right)}^{2}}} \right)$ because it's differentiation will give $\left( \dfrac{2}{{{\left( 1+t \right)}^{3}}} \right)$ i.e.
\[\begin{align}
  & \dfrac{d}{dt}\left( \dfrac{-1}{{{\left( 1+t \right)}^{2}}} \right)=-\dfrac{d}{dt}\left( {{\left( 1+t \right)}^{-2}} \right) \\
 & \Rightarrow -\left( -2 \right){{\left( 1+t \right)}^{-2-1}} \\
 & \Rightarrow 2{{\left( 1+t \right)}^{-3}} \\
 & \Rightarrow \left( \dfrac{2}{{{\left( 1+t \right)}^{-3}}} \right) \\
\end{align}\]
Hence,
\[\begin{align}
  & I=2\left( {{e}^{t}}\left( \dfrac{-1}{{{\left( 1+t \right)}^{2}}} \right) \right)+c \\
 & I=\dfrac{-2{{e}^{{{\tan }^{2}}x}}}{{{\left( 1+{{\tan }^{2}}x \right)}^{2}}}+c \\
\end{align}\]
Therefore, this is the required answer.

Note:
 During substitution like in this particular question ${{\tan }^{2}}x=t$ then we have to convert each term of x in the integral to term in t. Also, during implementing \[\int{{{e}^{x}}}\left( f\left( x \right)+f'\left( x \right) \right)dx={{e}^{x}}f\left( x \right)+c\] formula, always remember, there is addition of function and its derivative. If there is some other sign, then manipulate it to addition form. Otherwise, we will get wrong value of f(x) and final answer will be wrong.