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 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
$2\int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-2}}+{\frac {\cos(2x)e^{-2x}}{-2}}$.
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:
${\frac {e^{-2x}}{-4}}((\sin(2x)+\cos(2x))+C$

Return to Sample Exam