# 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

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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



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