1. **Problem Statement:** Calculate the integral $$\int e^x \cos x \, dx$$ using integration by parts. 2. **Formula Used:** Integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ 3. **Step 1:** Choose: - $$u = e^x$$ - $$dv = \cos x \, dx$$ Then, - $$du = e^x \, dx$$ - $$v = \sin x$$ 4. **Step 2:** Apply integration by parts: $$\int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx$$ 5. **Step 3:** Now evaluate $$\int e^x \sin x \, dx$$ using integration by parts again. Choose: - $$u = e^x$$ - $$dv = \sin x \, dx$$ Then, - $$du = e^x \, dx$$ - $$v = -\cos x$$ 6. **Step 4:** Apply integration by parts again: $$\int e^x \sin x \, dx = -e^x \cos x + \int e^x \cos x \, dx$$ 7. **Step 5:** Substitute back into the original integral: $$\int e^x \cos x \, dx = e^x \sin x - \left(-e^x \cos x + \int e^x \cos x \, dx\right) = e^x \sin x + e^x \cos x - \int e^x \cos x \, dx$$ 8. **Step 6:** Add $$\int e^x \cos x \, dx$$ to both sides: $$2 \int e^x \cos x \, dx = e^x (\sin x + \cos x)$$ 9. **Step 7:** Solve for the integral: $$\int e^x \cos x \, dx = \frac{e^x}{2} (\sin x + \cos x) + C$$ **Final Answer:** $$\boxed{\int e^x \cos x \, dx = \frac{e^x}{2} (\sin x + \cos x) + C}$$ --- Note: The integral is a classic example of repeated integration by parts leading to an equation involving the original integral, which can then be solved algebraically.