# 009B Sample Midterm 2, Problem 4 Detailed Solution

Evaluate the integral:

$\int e^{-2x}\sin(2x)~dx$ Background Information:
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
$\int e^{-2x}\sin(2x)~dx={\frac {\sin(2x)e^{-2x}}{-4}}+{\frac {\cos(2x)e^{-2x}}{-4}}.$ $\int e^{-2x}\sin(2x)~dx={\frac {e^{-2x}}{-4}}(\sin(2x)+\cos(2x))+C.$ ${\frac {e^{-2x}}{-4}}(\sin(2x)+\cos(2x))+C$ 