1708\hspace{-20}$Given $\bf{\int \sin (4x)\cdot e^{\tan^2 x}dx = 2\int \sin (2x)\cdot \cos (2x)\cdot e^{\tan^2 x}dx}$\\\\\\ Now Using $\bf{\bullet \; \sin (2x)=\frac{2\tan x}{1+\tan^2 x}}$ and $\bf{\bullet\; \cos(2x)=\frac{1-\tan^2 x}{1+\tan^2 x}}$\\\\\\ So Integral is $\bf{=4\int\frac{\tan x\cdot (1-\tan^2 x)}{(1+\tan^2 x)^2}\cdot e^{\tan^2 x}dx}$\\\\\\ Now Let $\bf{\tan^2 x= t\;,}$ Then $\bf{2\tan x\cdot \sec^2 xdx = dt\Rightarrow 2\tan xdx = \frac{1}{1+t^2}dt}$\\\\\\ So Integral is $\bf{=2\int\frac{\left(1-t\right)}{\left(1+t\right)^3}\cdot e^{t}dt = -2\int\left\{\frac{(t+1)-2}{(1+t)^3}\right\}\cdot e^{t}dt}$\\\\\\ So Integral is $\bf{=-2\int \left\{\frac{1}{(1+t)^2}-\frac{2}{(1+t)^3}\right\}\cdot e^tdt}$\\\\\\ Now Integral is Convert into $\bf{\int \left\{f(t)+f\hspace{-10}\quad '(t)\right\}\cdot e^{t}dt = e^t\cdot f(t)+\mathbb{C}}$\\\\\\
\hspace{-20}$So Original Integral Answer is $\bf{ = -\frac{2e^{t}}{(1+t)^2}+\mathbb{C}}$\\\\\\ So $\bf{\int \sin (4x)\cdot e^{\tan^2 x}dx = -\frac{2 e^{\tan^2 x}}{\left(1+\tan^{2}x\right)^2}+\mathbb{C}=-2\cos^4 x\cdot e^{\tan^2 x}+\mathbb{C}}$
