Evaluate the indefinite integral.
$$\int 2e^{2x}\sin \left(e^{2x}\right)dx=$$

Step 1 ：First, we recognize that this integral is a good candidate for integration by parts, which is given by the formula: $\int udv=uv-\int vdu$.

Step 2 ：We choose $u=e^{2x}$ and $dv=2e^{2x}\sin (e^{2x})dx$. Then we need to calculate $du$ and $v$.

Step 3 ：The derivative of $u=e^{2x}$ with respect to $x$ is $du=2e^{2x}dx$.

Step 4 ：To find $v$, we integrate $dv=2e^{2x}\sin (e^{2x})dx$. This is a standard integral of the form $\int e^{ax}\sin (bx)dx$, which equals $\frac{e^{ax}}{a^2+b^2}(a\sin (bx)-b\cos (bx))+C$. Here, $a=b=e^{2x}$, so $v=\frac{e^{2x}}{2e^{4x}+e^{4x}}(2e^{2x}\sin (e^{2x})-e^{2x}\cos (e^{2x}))=-\cos (e^{2x})+C$.

Step 5 ：Now we substitute $u$, $v$, $du$, and $dv$ into the integration by parts formula: $\int udv=uv-\int vdu=e^{2x}(-\cos (e^{2x}))-\int (-\cos (e^{2x}))(2e^{2x}dx)$.

Step 6 ：Simplify the integral to get: $-e^{2x}\cos (e^{2x})+2\int e^{2x}\cos (e^{2x})dx$.

Step 7 ：The integral $\int e^{2x}\cos (e^{2x})dx$ is a standard integral of the form $\int e^{ax}\cos (bx)dx$, which equals $\frac{e^{ax}}{a^2+b^2}(a\cos (bx)+b\sin (bx))+C$. Here, $a=b=e^{2x}$, so this integral equals $\frac{e^{2x}}{2e^{4x}+e^{4x}}(2e^{2x}\cos (e^{2x})+e^{2x}\sin (e^{2x}))=\sin (e^{2x})+C$.

Step 8 ：Substitute this result back into the equation to get the final result: $-e^{2x}\cos (e^{2x})+2\sin (e^{2x})+C$.

Step 9 ：Therefore, the indefinite integral $\int 2e^{2x}\sin \left(e^{2x}\right)dx=-e^{2x}\cos (e^{2x})+2\sin (e^{2x})+C$.

Step 10 ：This is the simplest form of the result, and it satisfies the requirements of the problem. Therefore, the final answer is $\boxed{{\displaystyle -e^{2x}\cos (e^{2x})+2\sin (e^{2x})+C}}$.