009B Sample Midterm 2, Problem 4

Evaluate the integral:

$\int e^{-2x}\sin(2x)~dx$

Foundations:
Review integration by parts

Solution:

Step 1:
We proceed using integration by parts. Let $u=\sin(2x)$ and $dv=e^{-2x}dx$. Then, $du=2\cos(2x)dx$ and $v={\frac {e^{-2x}}{-2}}$.
So, we get
$\int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-2}}-\int {\frac {e^{-2x}2\cos(2x)~dx}{-2}}={\frac {\sin(2x)e^{-2x}}{-2}}+\int e^{-2x}\cos(2x)~dx$

Step 2:
Now, we need to use integration by parts again. Let $u=\cos(2x)$ and $dv=e^{-2x}dx$. Then, $du=-2\sin(2x)dx$ and $v={\frac {e^{-2x}}{-2}}$.
So, we get
$\int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-2}}+{\frac {\cos(2x)e^{-2x}}{-2}}-\int e^{-2x}\sin(2x)~dx$.

Step 3:
Notice that the integral on the right of the last equation is the same integral that we had at the beginning.
So, if we add the integral on the right to the other side of the equation, we get
$2\int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-2}}+{\frac {\cos(2x)e^{-2x}}{-2}}$.
Now, we divide both sides by 2 to get
$\int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-4}}+{\frac {\cos(2x)e^{-2x}}{-4}}$.
Thus, the final answer is $\int e^{-2x}\sin(2x)~dx={\frac {e^{-2x}}{-4}}((\sin(2x)+\cos(2x))+C$

Final Answer:
${\frac {e^{-2x}}{-4}}((\sin(2x)+\cos(2x))+C$

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