1. **State the problem:** We need to evaluate the integral $$\int \cos x \ e^x \, dx$$. 2. **Choose method:** Use integration by parts, where we let: - $$u = \cos x$$ so that $$du = -\sin x \, dx$$ - $$dv = e^x \, dx$$ so that $$v = e^x$$ 3. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ Substitute the chosen parts: $$\int \cos x \, e^x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$$ 4. **Evaluate the new integral:** $$\int e^x \sin x \, dx$$ using integration by parts again. Let: - $$u = \sin x$$ so $$du = \cos x \, dx$$ - $$dv = e^x \, dx$$ so $$v = e^x$$ 5. **Apply integration by parts again:** $$\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx$$ 6. **Substitute back:** From step 3, we have: $$\int \cos x \, e^x \, dx = e^x \cos x + e^x \sin x - \int e^x \cos x \, dx$$ 7. **Solve for the integral:** Let $$I = \int e^x \cos x \, dx$$, then: $$I = e^x \cos x + e^x \sin x - I$$ 8. **Add $$I$$ to both sides:** $$2I = e^x (\cos x + \sin x)$$ 9. **Divide both sides by 2:** $$I = \frac{e^x (\cos x + \sin x)}{2} + C$$ **Final answer:** $$\boxed{\int \cos x \ e^x \, dx = \frac{e^x (\cos x + \sin x)}{2} + C}$$