Question

In Exercises \(17-20,\) use parts and solve for the unknown integral. $$\int e^{-x} \cos x d x$$

Short answer

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

Step-by-step solution

Choosing u and dv

In integration by parts, an integral of the form \(\int u dv\) becomes \(uv-\int v du\). For this exercise, let's choose \(u = e^{-x}\) and \(dv = \cos(x) dx\). The choice of 'u' and 'dv' is essential, and often, the selection is made based on which part of the integrand we can easily integrate or differentiate.

Calculate v and du

Now we need to calculate 'v' from 'dv' and 'du' from 'u'. The integral \(v\) of \(dv = \cos(x) dx\) is \(\sin(x)\), and the derivative 'du' of \(u = e^{-x}\) is \(-e^{-x} dx\).

Apply Integration by Parts

Substitute u, v, du into the formula for integration by parts \(uv-\int v du\). This results in \( e^{-x} \sin(x) - \int -e^{-x} \sin(x) dx \).

Solve for the Unknown Integral (part 1)

Observe that the result obtained in the previous step has another integral inside. This inner integral is similar to the original integral, but it is with \(\sin(x)\) instead of \(\cos(x)\). Let's solve it by once again applying integration by parts. Taking \(u1 = e^{-x}\) and \(dv1 = \sin(x) dx\), compute 'v1' from 'dv1' and 'du1' from 'u1'. The integral \(v1\) of \(dv1 = \sin(x) dx\) is \(-\cos(x)\), and the derivative 'du1' of \(u1 = e^{-x}\) is \(-e^{-x} dx\). Substitute back into the integration by parts formula: \( e^{-x} - \cos(x) - \int -e^{-x} - \cos(x) dx \).

Solve for the Unknown Integral (part 2)

Now that the integrals inside are a negative of the original integral, move it to the left side of the equation and solve for the original integral. This results in \(2 \int e^{-x} \cos(x) dx = e^{-x} \sin(x) + e^{-x} \cos(x) + C\). Therefore, the answer is \(\int e^{-x} \cos(x) dx = \frac{1}{2} (e^{-x} \sin(x) + e^{-x} \cos(x)) + \frac{C}{2}\).