To integrate functions involving $e^x$, $\cos x$, and $\sin x$, use integration by parts and simplify the result step by step. Don't forget the constant of integration, $C$.
We are given the integral:
\[
\int e^x (\cos x - \sin x) \, dx.
\]
To solve this, we can split the integral into two parts:
\[
\int e^x \cos x \, dx - \int e^x \sin x \, dx.
\]
Now, we can solve each of these integrals using integration by parts. Recall the formula for integration by parts:
\[
\int u \, dv = uv - \int v \, du.
\]
First Integral: $\int e^x \cos x \, dx$
Let:
- $u = \cos x \quad \text{and} \quad dv = e^x \, dx$,
so that:
- $du = -\sin x \, dx \quad \text{and} \quad v = e^x$.
Using the integration by parts formula:
\[
\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx.
\]
Second Integral: $\int e^x \sin x \, dx$
Let:
- $u = \sin x \quad \text{and} \quad dv = e^x \, dx$,
so that:
- $du = \cos x \, dx \quad \text{and} \quad v = e^x$.
Using the integration by parts formula:
\[
\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx.
\]
Substituting Back
Now substitute the second integral into the expression for the first integral:
\[
\int e^x \cos x \, dx = e^x \cos x + \left( e^x \sin x - \int e^x \cos x \, dx \right).
\]
Simplifying:
\[
\int e^x \cos x \, dx + \int e^x \cos x \, dx = e^x \cos x + e^x \sin x,
\]
\[
2 \int e^x \cos x \, dx = e^x (\cos x + \sin x),
\]
\[
\int e^x \cos x \, dx = \frac{e^x (\cos x + \sin x)}{2}.
\]
Thus, the final integral is:
\[
\int e^x (\cos x - \sin x) \, dx = e^x \cos x + C.
\]
