Integral of $$$e^{9 x} \cos{\left(x \right)}$$$

Find $$$\int e^{9 x} \cos{\left(x \right)}\, dx$$$.

For the integral $$$\int{e^{9 x} \cos{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\cos{\left(x \right)}$$$ and $$$\operatorname{dv}=e^{9 x} dx$$$.

Then $$$\operatorname{du}=\left(\cos{\left(x \right)}\right)^{\prime }dx=- \sin{\left(x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{9 x} d x}=\frac{e^{9 x}}{9}$$$ (steps can be seen »).

Thus,

$${\color{red}{\int{e^{9 x} \cos{\left(x \right)} d x}}}={\color{red}{\left(\cos{\left(x \right)} \cdot \frac{e^{9 x}}{9}-\int{\frac{e^{9 x}}{9} \cdot \left(- \sin{\left(x \right)}\right) d x}\right)}}={\color{red}{\left(\frac{e^{9 x} \cos{\left(x \right)}}{9} - \int{\left(- \frac{e^{9 x} \sin{\left(x \right)}}{9}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{9}$$$ and $$$f{\left(x \right)} = e^{9 x} \sin{\left(x \right)}$$$:

$$\frac{e^{9 x} \cos{\left(x \right)}}{9} - {\color{red}{\int{\left(- \frac{e^{9 x} \sin{\left(x \right)}}{9}\right)d x}}} = \frac{e^{9 x} \cos{\left(x \right)}}{9} - {\color{red}{\left(- \frac{\int{e^{9 x} \sin{\left(x \right)} d x}}{9}\right)}}$$

For the integral $$$\int{e^{9 x} \sin{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\sin{\left(x \right)}$$$ and $$$\operatorname{dv}=e^{9 x} dx$$$.

Then $$$\operatorname{du}=\left(\sin{\left(x \right)}\right)^{\prime }dx=\cos{\left(x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{9 x} d x}=\frac{e^{9 x}}{9}$$$ (steps can be seen »).

Thus,

$$\frac{e^{9 x} \cos{\left(x \right)}}{9} + \frac{{\color{red}{\int{e^{9 x} \sin{\left(x \right)} d x}}}}{9}=\frac{e^{9 x} \cos{\left(x \right)}}{9} + \frac{{\color{red}{\left(\sin{\left(x \right)} \cdot \frac{e^{9 x}}{9}-\int{\frac{e^{9 x}}{9} \cdot \cos{\left(x \right)} d x}\right)}}}{9}=\frac{e^{9 x} \cos{\left(x \right)}}{9} + \frac{{\color{red}{\left(\frac{e^{9 x} \sin{\left(x \right)}}{9} - \int{\frac{e^{9 x} \cos{\left(x \right)}}{9} d x}\right)}}}{9}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{9}$$$ and $$$f{\left(x \right)} = e^{9 x} \cos{\left(x \right)}$$$:

$$\frac{e^{9 x} \sin{\left(x \right)}}{81} + \frac{e^{9 x} \cos{\left(x \right)}}{9} - \frac{{\color{red}{\int{\frac{e^{9 x} \cos{\left(x \right)}}{9} d x}}}}{9} = \frac{e^{9 x} \sin{\left(x \right)}}{81} + \frac{e^{9 x} \cos{\left(x \right)}}{9} - \frac{{\color{red}{\left(\frac{\int{e^{9 x} \cos{\left(x \right)} d x}}{9}\right)}}}{9}$$

We've arrived to an integral that we already saw.

Thus, we've obtained the following simple equation with respect to the integral:

$$\int{e^{9 x} \cos{\left(x \right)} d x} = \frac{e^{9 x} \sin{\left(x \right)}}{81} + \frac{e^{9 x} \cos{\left(x \right)}}{9} - \frac{\int{e^{9 x} \cos{\left(x \right)} d x}}{81}$$

Solving it, we get that

$$\int{e^{9 x} \cos{\left(x \right)} d x} = \frac{\left(\sin{\left(x \right)} + 9 \cos{\left(x \right)}\right) e^{9 x}}{82}$$

Therefore,

$$\int{e^{9 x} \cos{\left(x \right)} d x} = \frac{\left(\sin{\left(x \right)} + 9 \cos{\left(x \right)}\right) e^{9 x}}{82}$$

Add the constant of integration:

$$\int{e^{9 x} \cos{\left(x \right)} d x} = \frac{\left(\sin{\left(x \right)} + 9 \cos{\left(x \right)}\right) e^{9 x}}{82}+C$$

$$$\int e^{9 x} \cos{\left(x \right)}\, dx = \frac{\left(\sin{\left(x \right)} + 9 \cos{\left(x \right)}\right) e^{9 x}}{82} + C$$$A