1. **State the problem:** We need to find the indefinite integral $$\int \left(e^{2x} - 5 \sin 2x\right) \, dx$$. 2. **Recall the integral formulas:** - The integral of an exponential function $$\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C$$. - The integral of sine function $$\int \sin bx \, dx = -\frac{1}{b} \cos bx + C$$. 3. **Apply linearity of integration:** $$\int \left(e^{2x} - 5 \sin 2x\right) \, dx = \int e^{2x} \, dx - 5 \int \sin 2x \, dx$$. 4. **Integrate each term:** - For $$\int e^{2x} \, dx$$, using the formula, $$a=2$$: $$\int e^{2x} \, dx = \frac{1}{2} e^{2x} + C_1$$. - For $$\int \sin 2x \, dx$$, using the formula, $$b=2$$: $$\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C_2$$. 5. **Substitute back and simplify:** $$\int \left(e^{2x} - 5 \sin 2x\right) \, dx = \frac{1}{2} e^{2x} - 5 \left(-\frac{1}{2} \cos 2x\right) + C$$ 6. **Simplify the expression:** $$= \frac{1}{2} e^{2x} + \frac{5}{2} \cos 2x + C$$ **Final answer:** $$\boxed{\int \left(e^{2x} - 5 \sin 2x\right) \, dx = \frac{1}{2} e^{2x} + \frac{5}{2} \cos 2x + C}$$