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 can 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,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=e^{-2x}dx.}
Then,  $du=2\cos(2x)dx$  and  $v={\frac {e^{-2x}}{-2}}.$
Thus, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=e^{-2x}dx.}
Then,  $du=-2\sin(2x)dx$  and  $v={\frac {e^{-2x}}{-2}}.$
Therefore, 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.
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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