How to Solve the Integral of e^-x cos(2x)

Solving the Integral of $ e^{-x} \cos(2x) $

Integrating functions involving exponential and trigonometric terms can often seem daunting. However, with the right approach, these integrals can be managed efficiently. This article will guide you through the process of solving $ \int e^{-x} \cos(2x) \, dx $, a common integral in calculus, using integration by parts.

Setting Up the Integral

To solve $ \int e^{-x} \cos(2x) \, dx $, we start by applying the integration by parts formula. Integration by parts is based on the product rule for differentiation and is given by:

$$ \int u \, dv = uv - \int v \, du $$

We choose our parts as follows:

$ u = e^{-x} $ (this makes $ du $ simpler)
$ dv = \cos(2x) \, dx $ (which integrates easily)

Then, we calculate $ du $ and $ v $:

$ du = -e^{-x} \, dx $
$ v = \frac{1}{2} \sin(2x) $ (integration of $ \cos(2x) $)

Applying Integration by Parts

Substituting the values we calculated into the integration by parts formula gives:

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

This simplifies to:

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

Reapplying Integration by Parts

To solve $ \int e^{-x} \sin(2x) \, dx $, we use integration by parts again with:

$ u = e^{-x} $ and $ dv = \sin(2x) \, dx $

Proceeding as before, we find:

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

Final Calculation

These steps lead to a system of equations for $ \int e^{-x} \cos(2x) \, dx $ and $ \int e^{-x} \sin(2x) \, dx $. Solving these gives:

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

This result shows the antiderivative of the original function.

Conclusion