Integral of cos(x)*exp(2x+7) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$e^{2 x + 7} \cos{\left(x \right)} = e^{7} e^{2 x} \cos{\left(x \right)}$$

   2. The integral of a constant times a function is the constant times the integral of the function:

      $$\int e^{7} e^{2 x} \cos{\left(x \right)}\, dx = e^{7} \int e^{2 x} \cos{\left(x \right)}\, dx$$

      1. Use integration by parts, noting that the integrand eventually repeats itself.

         1. For the integrand $e^{2 x} \cos{\left(x \right)}$:

            Let $u{\left(x \right)} = \cos{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

            Then $\int e^{2 x} \cos{\left(x \right)}\, dx = \frac{e^{2 x} \cos{\left(x \right)}}{2} - \int \left(- \frac{e^{2 x} \sin{\left(x \right)}}{2}\right)\, dx$.

         2. For the integrand $- \frac{e^{2 x} \sin{\left(x \right)}}{2}$:

            Let $u{\left(x \right)} = - \frac{\sin{\left(x \right)}}{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

            Then $\int e^{2 x} \cos{\left(x \right)}\, dx = \frac{e^{2 x} \sin{\left(x \right)}}{4} + \frac{e^{2 x} \cos{\left(x \right)}}{2} + \int \left(- \frac{e^{2 x} \cos{\left(x \right)}}{4}\right)\, dx$.

         3. Notice that the integrand has repeated itself, so move it to one side:

            $$\frac{5 \int e^{2 x} \cos{\left(x \right)}\, dx}{4} = \frac{e^{2 x} \sin{\left(x \right)}}{4} + \frac{e^{2 x} \cos{\left(x \right)}}{2}$$

            Therefore,

            $$\int e^{2 x} \cos{\left(x \right)}\, dx = \frac{e^{2 x} \sin{\left(x \right)}}{5} + \frac{2 e^{2 x} \cos{\left(x \right)}}{5}$$

      So, the result is: $\left(\frac{e^{2 x} \sin{\left(x \right)}}{5} + \frac{2 e^{2 x} \cos{\left(x \right)}}{5}\right) e^{7}$

   Method #2

   1. Rewrite the integrand:

      $$e^{2 x + 7} \cos{\left(x \right)} = e^{7} e^{2 x} \cos{\left(x \right)}$$

   2. The integral of a constant times a function is the constant times the integral of the function:

      $$\int e^{7} e^{2 x} \cos{\left(x \right)}\, dx = e^{7} \int e^{2 x} \cos{\left(x \right)}\, dx$$

      1. Use integration by parts, noting that the integrand eventually repeats itself.

         1. For the integrand $e^{2 x} \cos{\left(x \right)}$:

            Let $u{\left(x \right)} = \cos{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

            Then $\int e^{2 x} \cos{\left(x \right)}\, dx = \frac{e^{2 x} \cos{\left(x \right)}}{2} - \int \left(- \frac{e^{2 x} \sin{\left(x \right)}}{2}\right)\, dx$.

         2. For the integrand $- \frac{e^{2 x} \sin{\left(x \right)}}{2}$:

            Let $u{\left(x \right)} = - \frac{\sin{\left(x \right)}}{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

            Then $\int e^{2 x} \cos{\left(x \right)}\, dx = \frac{e^{2 x} \sin{\left(x \right)}}{4} + \frac{e^{2 x} \cos{\left(x \right)}}{2} + \int \left(- \frac{e^{2 x} \cos{\left(x \right)}}{4}\right)\, dx$.

         3. Notice that the integrand has repeated itself, so move it to one side:

            $$\frac{5 \int e^{2 x} \cos{\left(x \right)}\, dx}{4} = \frac{e^{2 x} \sin{\left(x \right)}}{4} + \frac{e^{2 x} \cos{\left(x \right)}}{2}$$

            Therefore,

            $$\int e^{2 x} \cos{\left(x \right)}\, dx = \frac{e^{2 x} \sin{\left(x \right)}}{5} + \frac{2 e^{2 x} \cos{\left(x \right)}}{5}$$

      So, the result is: $\left(\frac{e^{2 x} \sin{\left(x \right)}}{5} + \frac{2 e^{2 x} \cos{\left(x \right)}}{5}\right) e^{7}$

2. Now simplify:

   $$\frac{\left(\sin{\left(x \right)} + 2 \cos{\left(x \right)}\right) e^{2 x + 7}}{5}$$

3. Add the constant of integration:

   $$\frac{\left(\sin{\left(x \right)} + 2 \cos{\left(x \right)}\right) e^{2 x + 7}}{5}+ \mathrm{constant}$$

The answer is: