To integrate $\int e^{-2x} \sin(x) \, dx$ using integration by parts, we can proceed as follows:

Step 1: Set Up Integration by Parts

For integration by parts, we use the formula: $\int u \, dv = uv - \int v \, du$

Choose:

• $u = \sin(x)$ (so that $du = \cos(x) \, dx$)
• $dv = e^{-2x} \, dx$ (so that $v = \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}$)

Step 2: Apply Integration by Parts

Substitute $u$ and $dv$ into the integration by parts formula: $\int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} - \int -\frac{1}{2} e^{-2x} \cos(x) \, dx$

This simplifies to: $\int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} + \frac{1}{2} \int e^{-2x} \cos(x) \, dx$

Step 3: Integrate $\int e^{-2x} \cos(x) \, dx$ by Parts Again

Now, we need to integrate $\int e^{-2x} \cos(x) \, dx$ using integration by parts again.

Choose:

• $u = \cos(x)$ (so that $du = -\sin(x) \, dx$)
• $dv = e^{-2x} \, dx$ (so that $v = -\frac{1}{2} e^{-2x}$)

Applying integration by parts to $\int e^{-2x} \cos(x) \, dx$: $\int e^{-2x} \cos(x) \, dx = -\frac{1}{2} \cos(x) e^{-2x} - \int -\frac{1}{2} e^{-2x} (-\sin(x)) \, dx$

Simplify: $\int e^{-2x} \cos(x) \, dx = -\frac{1}{2} \cos(x) e^{-2x} - \frac{1}{2} \int e^{-2x} \sin(x) \, dx$

Step 4: Set Up a System of Equations

Now we have the original integral $\int e^{-2x} \sin(x) \, dx$ in terms of itself: $\int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} + \frac{1}{2} \left(-\frac{1}{2} \cos(x) e^{-2x} - \frac{1}{2} \int e^{-2x} \sin(x) \, dx \right)$

Simplify and solve for $\int e^{-2x} \sin(x) \, dx$: $\int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} - \frac{1}{4} \cos(x) e^{-2x} - \frac{1}{4} \int e^{-2x} \sin(x) \, dx$

Combine terms to isolate the integral: $\int e^{-2x} \sin(x) \, dx + \frac{1}{4} \int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} - \frac{1}{4} \cos(x) e^{-2x}$

Factor out $\int e^{-2x} \sin(x) \, dx$: $\left(1 + \frac{1}{4}\right) \int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} - \frac{1}{4} \cos(x) e^{-2x}$

This simplifies to: $\frac{5}{4} \int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} - \frac{1}{4} \cos(x) e^{-2x}$

Finally, solve for $\int e^{-2x} \sin(x) \, dx$: $\int e^{-2x} \sin(x) \, dx = -\frac{1}{2} \sin(x) e^{-2x} - \frac{1}{4} \cos(x) e^{-2x} \cdot \frac{4}{5}$

So, $\int e^{-2x} \sin(x) \, dx = -\frac{2}{5} \sin(x) e^{-2x} - \frac{1}{5} \cos(x) e^{-2x} + C$

Let me know if you'd like more clarification or have further questions.

1. How do we choose functions $u$ and $dv$ in integration by parts?
2. Can we always use integration by parts to solve integrals like this?
3. What are other methods for solving complex integrals?
4. How would we solve a similar integral involving $e^{ax} \cos(bx)$?
5. How do we apply integration by parts in cases with higher powers of $x$ alongside exponential functions?

Tip: When using integration by parts twice leads to the original integral, setting up a system of equations to solve for it is a powerful trick.