009B Sample Midterm 2, Problem 4

Evaluate the integral:

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

Foundations:
1. Integration by parts tells us $\int u~dv=uv-\int v~du.$
2. How would you integrate $\int e^{x}\sin x~dx?$
You could use integration by parts.
Let $u=\sin(x)$ and $dv=e^{x}~dx.$ Then, $du=\cos(x)~dx$ and $v=e^{x}.$
Thus, $\int e^{x}\sin x~dx=e^{x}\sin(x)-\int e^{x}\cos(x)~dx.$
Now, we need to use integration by parts a second time.
Let $u=\cos(x)$ and $dv=e^{x}~dx.$ Then, $du=-\sin(x)~dx$ and $v=e^{x}.$ So,
${\begin{array}{rcl}\displaystyle {\int e^{x}\sin x~dx}&=&\displaystyle {e^{x}\sin(x)-(e^{x}\cos(x)-\int -e^{x}\sin(x)~dx}\\&&\\&=&\displaystyle {e^{x}(\sin(x)-\cos(x))-\int e^{x}\sin(x)~dx.}\\\end{array}}$
Notice, we are back where we started. So, adding the last term on the right hand side to the opposite side,
we get $2\int e^{x}\sin(x)~dx=e^{x}(\sin(x)-\cos(x)).$
Hence, $\int e^{x}\sin(x)~dx={\frac {e^{x}}{2}}(\sin(x)-\cos(x))+C.$

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

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