`int e^(-theta) cos(2 theta) d theta` Evaluate the integral

We have to find the integral `\int e^{-\theta}cos(2\theta) d\theta`

We can do this by integration by parts i.e.

`\int e^{-\theta}cos(2\theta) d\theta=e^{-\theta}\int cos(2\theta) d\theta -\int( \frac{d}{d\theta}(e^{-\theta})\int cos(2\theta) d\theta )d\theta`

                       `=e^{-\theta}.\frac{sin(2\theta)}{2}-\int-e^{-\theta}.\frac{sin(2\theta)}{2} d\theta`

                        `=e^{-\theta}.\frac{sin(2\theta)}{2}+\frac{1}{2}\int e^{-\theta}sin(2\theta) d\theta`

                         `=e^{-\theta}.\frac{sin(2\theta)}{2}+\frac{1}{2}[e^{-\theta}\int sin(2\theta) d\theta-\int -e^{-\theta}.\int sin(2\theta) d\theta]`

                          `=\frac{e^{-\theta}sin(2\theta)}{2}+\frac{1}{2}[\frac{-e^{-\theta}cos(2\theta)}{2}-\frac{1}{2}\int e^{-\theta}cos(2\theta )d\theta]`

                          `=\frac{e^{-\theta}sin(2\theta)}{2}-\frac{e^{-\theta}cos(2\theta)}{4}-\frac{1}{4}\int e^{-\theta}cos(2\theta) d\theta`

i.e. `\frac{5}{4}\int e^{-\theta}cos(2\theta)d\theta= \frac{e^{-\theta}(2sin(2\theta)-cos(2\theta))}{4}`

i.e. `\int e^{-\theta }cos(2\theta) d\theta= \frac{e^{-\theta}}{5} (2sin(2\theta)-cos(2\theta)) +C`

``

We have to find the integral `\int e^{-\theta}cos(2\theta) d\theta`

We can do this by integration by parts i.e.

`\int e^{-\theta}cos(2\theta) d\theta=e^{-\theta}\int cos(2\theta) d\theta -\int( \frac{d}{d\theta}(e^{-\theta})\int cos(2\theta) d\theta )d\theta`

                       `=e^{-\theta}.\frac{sin(2\theta)}{2}-\int-e^{-\theta}.\frac{sin(2\theta)}{2} d\theta`

                        `=e^{-\theta}.\frac{sin(2\theta)}{2}+\frac{1}{2}\int e^{-\theta}sin(2\theta) d\theta`

                         `=e^{-\theta}.\frac{sin(2\theta)}{2}+\frac{1}{2}[e^{-\theta}\int sin(2\theta) d\theta-\int -e^{-\theta}.\int sin(2\theta) d\theta]`

                          `=\frac{e^{-\theta}sin(2\theta)}{2}+\frac{1}{2}[\frac{-e^{-\theta}cos(2\theta)}{2}-\frac{1}{2}\int e^{-\theta}cos(2\theta )d\theta]`

                          `=\frac{e^{-\theta}sin(2\theta)}{2}-\frac{e^{-\theta}cos(2\theta)}{4}-\frac{1}{4}\int e^{-\theta}cos(2\theta) d\theta`

i.e. `\frac{5}{4}\int e^{-\theta}cos(2\theta)d\theta= \frac{e^{-\theta}(2sin(2\theta)-cos(2\theta))}{4}`

i.e. `\int e^{-\theta }cos(2\theta) d\theta= \frac{e^{-\theta}}{5} (2sin(2\theta)-cos(2\theta)) +C`

``