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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 in Step 2 is the same integral that we had at the beginning of the problem.
So, if we add the integral on the right to the other side of the equation, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-4}}+{\frac {\cos(2x)e^{-2x}}{-4}}}  .
Thus, the final answer is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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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