1. **Problem statement:** Evaluate the integral $$\int \frac{1}{r} e^{rx} \cos x \, dx$$. 2. **Given substitutions:** - Let $$u = \cos x$$, so $$du = -\sin x \, dx$$. - Let $$dv = e^{rx} dx$$, so $$v = \int e^{rx} dx = \frac{1}{r} e^{rx}$$. 3. **Integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$. 4. **Apply integration by parts:** $$\int \frac{1}{r} e^{rx} \cos x \, dx = u v - \int v \, du = \cos x \cdot \frac{1}{r} e^{rx} - \int \frac{1}{r} e^{rx} (-\sin x) \, dx$$ 5. **Simplify the integral:** $$= \frac{1}{r} e^{rx} \cos x + \frac{1}{r} \int e^{rx} \sin x \, dx$$ 6. **Now evaluate $$\int e^{rx} \sin x \, dx$$ using integration by parts again:** - Let $$I = \int e^{rx} \sin x \, dx$$. 7. **Integration by parts for $$I$$:** - Let $$u = \sin x$$, so $$du = \cos x \, dx$$. - Let $$dv = e^{rx} dx$$, so $$v = \frac{1}{r} e^{rx}$$. 8. **Apply integration by parts:** $$I = u v - \int v \, du = \sin x \cdot \frac{1}{r} e^{rx} - \int \frac{1}{r} e^{rx} \cos x \, dx$$ 9. **Rewrite:** $$I = \frac{1}{r} e^{rx} \sin x - \frac{1}{r} \int e^{rx} \cos x \, dx$$ 10. **Notice the original integral appears again:** Let $$J = \int e^{rx} \cos x \, dx$$. 11. **From step 5, we have:** $$\int \frac{1}{r} e^{rx} \cos x \, dx = \frac{1}{r} J$$. 12. **Substitute $$I$$ and $$J$$ back:** From step 9: $$I = \frac{1}{r} e^{rx} \sin x - \frac{1}{r} J$$ From step 5: $$\frac{1}{r} J = \frac{1}{r} e^{rx} \cos x + \frac{1}{r} I$$ Multiply both sides by $$r$$: $$J = e^{rx} \cos x + I$$ 13. **Substitute $$J$$ into expression for $$I$$:** $$I = \frac{1}{r} e^{rx} \sin x - \frac{1}{r} J = \frac{1}{r} e^{rx} \sin x - \frac{1}{r} (e^{rx} \cos x + I)$$ 14. **Simplify:** $$I = \frac{1}{r} e^{rx} \sin x - \frac{1}{r} e^{rx} \cos x - \frac{1}{r} I$$ 15. **Group $$I$$ terms:** $$I + \frac{1}{r} I = \frac{1}{r} e^{rx} (\sin x - \cos x)$$ 16. **Factor out $$I$$:** $$I \left(1 + \frac{1}{r}\right) = \frac{1}{r} e^{rx} (\sin x - \cos x)$$ 17. **Simplify the factor:** $$I \frac{r+1}{r} = \frac{1}{r} e^{rx} (\sin x - \cos x)$$ 18. **Solve for $$I$$:** $$I = \frac{\frac{1}{r} e^{rx} (\sin x - \cos x)}{\frac{r+1}{r}} = \frac{e^{rx} (\sin x - \cos x)}{r+1}$$ 19. **Recall from step 5:** $$\int \frac{1}{r} e^{rx} \cos x \, dx = \frac{1}{r} J = \frac{1}{r} (e^{rx} \cos x + I)$$ 20. **Substitute $$I$$:** $$= \frac{1}{r} e^{rx} \cos x + \frac{1}{r} \cdot \frac{e^{rx} (\sin x - \cos x)}{r+1} = \frac{e^{rx}}{r} \cos x + \frac{e^{rx}}{r(r+1)} (\sin x - \cos x)$$ 21. **Combine terms:** $$= e^{rx} \left( \frac{\cos x}{r} + \frac{\sin x - \cos x}{r(r+1)} \right) = e^{rx} \left( \frac{\cos x (r+1) + \sin x - \cos x}{r(r+1)} \right)$$ 22. **Simplify numerator:** $$\cos x (r+1) - \cos x = r \cos x$$ So numerator is: $$r \cos x + \sin x$$ 23. **Final answer:** $$\int \frac{1}{r} e^{rx} \cos x \, dx = \frac{e^{rx}}{r(r+1)} (r \cos x + \sin x) + C$$