Problems on Techniques of Integration

Use the Integration by Parts. Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=& \sin(2x)\\
dv &=& e^{3x} dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=& 2 \cos(2x) dx\\
v &=& \displaystyle \frac{1}{3} e^{3x}\;.
\end{array}\right.\end{displaymath}

So

\begin{displaymath}\begin{array}{lll}
\displaystyle \int \sin(2x)e^{3x} dx &=&\d...
...in(2x)e^{3x} - \frac{2}{3}\int \cos(2x)e^{3x} dx
\end{array}\;.\end{displaymath}

In order to integrate the function $\cos(2x)e^{3x}$, we will need to do another integration by parts. Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=&\cos(2x)\\
dv &=& e^{3x}dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&-2 \sin(2x)dx\\
v &=&\displaystyle \frac{1}{3} e^{3x}\;.
\end{array}\right.\end{displaymath}

So

\begin{displaymath}\begin{array}{lll}
\displaystyle \int \cos(2x)e^{3x} dx &=&\d...
...os(2x)e^{3x} + \frac{2}{3}\int \sin(2x)e^{3x} dx
\end{array}\;.\end{displaymath}

Hence

\begin{displaymath}\int \sin(2x)e^{3x} dx = \frac{1}{3}\sin(2x)e^{3x} - \frac{2}...
...3}\cos(2x)e^{3x} + \frac{2}{3}\int \sin(2x)e^{3x} dx\right] \;.\end{displaymath}

which implies

\begin{displaymath}\left(1 + \frac{4}{9}\right) \int \sin(2 x)e^{3 x} dx = \frac{1}{3}\sin(2x)e^{3x} - \frac{2}{9}\cos(2x)e^{3x} \end{displaymath}

or

\begin{displaymath}\int \sin(2x)e^{3x} dx = \frac{3}{13} \sin(2x)e^{3x} - \frac{2}{13} \cos(2x)e^{3x} + C\;.\end{displaymath}

Detailed Answer.


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