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 $v=e^{x}.$
Therefore,
${\begin{array}{rcl}\displaystyle {\int e^{x}\sin x~dx}&=&\displaystyle {e^{x}\sin(x)-(e^{x}\cos(x)-\int -e^{x}\sin(x)~dx}\\&&\\&=&\displaystyle {e^{x}(\sin(x)-\cos(x))-\int e^{x}\sin(x)~dx.}\\\end{array}}$
Notice, we are back where we started.
Therefore, 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 $v={\frac {e^{-2x}}{-2}}.$
Thus, we get
${\begin{array}{rcl}\displaystyle {\int e^{-2x}\sin(2x)~dx}&=&\displaystyle {{\frac {\sin(2x)e^{-2x}}{-2}}-\int {\frac {e^{-2x}2\cos(2x)~dx}{-2}}}\\&&\\&=&\displaystyle {{\frac {\sin(2x)e^{-2x}}{-2}}+\int e^{-2x}\cos(2x)~dx.}\end{array}}$

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 $v={\frac {e^{-2x}}{-2}}.$
Therefore, 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.
Thus, if we add the integral on the right to the other side of the equation, we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^{-2x}\sin (2x)~dx=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C.}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C}

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