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