# 009B Sample Midterm 2, Problem 4 Detailed Solution

Evaluate the integral:

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

Background Information:
1. Integration by parts tells us
${\displaystyle \int u~dv=uv-\int v~du.}$
2. How would you integrate  ${\displaystyle \int e^{x}\sin x~dx?}$

You can use integration by parts.

Let  ${\displaystyle u=\sin(x)}$  and  ${\displaystyle dv=e^{x}~dx.}$

Then,  ${\displaystyle du=\cos(x)~dx}$  and  ${\displaystyle v=e^{x}.}$

Thus,  ${\displaystyle \int e^{x}\sin x~dx=e^{x}\sin(x)-\int e^{x}\cos(x)~dx.}$

Now, we need to use integration by parts a second time.

Let  ${\displaystyle u=\cos(x)}$  and  ${\displaystyle dv=e^{x}~dx.}$

Then,  ${\displaystyle du=-\sin(x)~dx}$  and  ${\displaystyle v=e^{x}.}$
Therefore,

${\displaystyle {\begin{array}{rcl}\displaystyle {\int e^{x}\sin x~dx}&=&\displaystyle {e^{x}\sin(x)-(e^{x}\cos(x)-\int -e^{x}\sin(x)~dx}\\&&\\&=&\displaystyle {e^{x}(\sin(x)-\cos(x))-\int e^{x}\sin(x)~dx.}\\\end{array}}}$

Notice, we are back where we started.

Therefore, adding the last term on the right hand side to the opposite side, we get

${\displaystyle 2\int e^{x}\sin(x)~dx=e^{x}(\sin(x)-\cos(x)).}$

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

Solution:

Step 1:
We proceed using integration by parts.
Let  ${\displaystyle u=\sin(2x)}$  and  ${\displaystyle dv=e^{-2x}dx.}$
Then,  ${\displaystyle du=2\cos(2x)dx}$  and  ${\displaystyle v={\frac {e^{-2x}}{-2}}.}$
Thus, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {\int e^{-2x}\sin(2x)~dx}&=&\displaystyle {{\frac {\sin(2x)e^{-2x}}{-2}}-\int {\frac {e^{-2x}2\cos(2x)~dx}{-2}}}\\&&\\&=&\displaystyle {{\frac {\sin(2x)e^{-2x}}{-2}}+\int e^{-2x}\cos(2x)~dx.}\end{array}}}$

Step 2:
Now, we need to use integration by parts again.
Let  ${\displaystyle u=\cos(2x)}$  and  ${\displaystyle dv=e^{-2x}dx.}$
Then,  ${\displaystyle du=-2\sin(2x)dx}$  and  ${\displaystyle v={\frac {e^{-2x}}{-2}}.}$
Therefore, we get

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

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
${\displaystyle 2\int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-2}}+{\frac {\cos(2x)e^{-2x}}{-2}}.}$
Now, we divide both sides by 2 to get
${\displaystyle \int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-4}}+{\frac {\cos(2x)e^{-2x}}{-4}}.}$
Thus, the final answer is
${\displaystyle \int e^{-2x}\sin(2x)~dx={\frac {e^{-2x}}{-4}}(\sin(2x)+\cos(2x))+C.}$

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