Evaluate the indefinite integral.
\[
\int 2 e^{2 x} \sin \left(e^{2 x}\right) d x=
\]

Step 1 ：First, we recognize that this integral is a good candidate for integration by parts, which is given by the formula: \(\int u dv = uv - \int v du\).

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: \(\int u dv = uv - \int v du = e^{2x} (-\cos(e^{2x})) - \int (-\cos(e^{2x})) (2e^{2x} dx)\).

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 \(\int 2 e^{2 x} \sin \left(e^{2 x}\right) d x = -e^{2x} \cos(e^{2x}) + 2 \sin(e^{2x}) + C\).

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