Integral of e^(2*x)*sin(2*x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{e^{u} \sin{\left(u \right)}}{4}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{e^{u} \sin{\left(u \right)}}{2}\, du = \frac{\int e^{u} \sin{\left(u \right)}\, du}{2}$$

         1. Use integration by parts, noting that the integrand eventually repeats itself.

            1. For the integrand $e^{u} \sin{\left(u \right)}$:

               Let $u{\left(u \right)} = \sin{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

               Then $\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - \int e^{u} \cos{\left(u \right)}\, du$.

            2. For the integrand $e^{u} \cos{\left(u \right)}$:

               Let $u{\left(u \right)} = \cos{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

               Then $\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)} + \int \left(- e^{u} \sin{\left(u \right)}\right)\, du$.

            3. Notice that the integrand has repeated itself, so move it to one side:

               $$2 \int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)}$$

               Therefore,

               $$\int e^{u} \sin{\left(u \right)}\, du = \frac{e^{u} \sin{\left(u \right)}}{2} - \frac{e^{u} \cos{\left(u \right)}}{2}$$

         So, the result is: $\frac{e^{u} \sin{\left(u \right)}}{4} - \frac{e^{u} \cos{\left(u \right)}}{4}$

      Now substitute $u$ back in:

      $$\frac{e^{2 x} \sin{\left(2 x \right)}}{4} - \frac{e^{2 x} \cos{\left(2 x \right)}}{4}$$

   Method #2

   1. Use integration by parts, noting that the integrand eventually repeats itself.

      1. For the integrand $e^{2 x} \sin{\left(2 x \right)}$:

         Let $u{\left(x \right)} = \sin{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

         Then $\int e^{2 x} \sin{\left(2 x \right)}\, dx = \frac{e^{2 x} \sin{\left(2 x \right)}}{2} - \int e^{2 x} \cos{\left(2 x \right)}\, dx$.

      2. For the integrand $e^{2 x} \cos{\left(2 x \right)}$:

         Let $u{\left(x \right)} = \cos{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

         Then $\int e^{2 x} \sin{\left(2 x \right)}\, dx = \frac{e^{2 x} \sin{\left(2 x \right)}}{2} - \frac{e^{2 x} \cos{\left(2 x \right)}}{2} + \int \left(- e^{2 x} \sin{\left(2 x \right)}\right)\, dx$.

      3. Notice that the integrand has repeated itself, so move it to one side:

         $$2 \int e^{2 x} \sin{\left(2 x \right)}\, dx = \frac{e^{2 x} \sin{\left(2 x \right)}}{2} - \frac{e^{2 x} \cos{\left(2 x \right)}}{2}$$

         Therefore,

         $$\int e^{2 x} \sin{\left(2 x \right)}\, dx = \frac{e^{2 x} \sin{\left(2 x \right)}}{4} - \frac{e^{2 x} \cos{\left(2 x \right)}}{4}$$

2. Now simplify:

   $$- \frac{\sqrt{2} e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)}}{4}$$

3. Add the constant of integration:

   $$- \frac{\sqrt{2} e^{2 x} \cos{\left(2 x + \frac{\pi}{4} \right)}}{4}+ \mathrm{constant}$$

The answer is: