Problems on Techniques of Integration

Use the Integration by Parts. Set

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

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=& -a \sin(ax) dx\\
v &=& \displaystyle \frac{1}{b} e^{bx}\;.
\end{array}\right.\end{displaymath}

So

\begin{displaymath}\begin{array}{lll}
\displaystyle \int \cos(ax)e^{bx} dx &=&\d...
...os(ax)e^{bx} + \frac{a}{b}\int \sin(ax)e^{bx} dx
\end{array}\;.\end{displaymath}

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

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

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&a \cos(ax)dx\\
v &=&\displaystyle \frac{1}{b} e^{bx}\;.
\end{array}\right.\end{displaymath}

So

\begin{displaymath}\begin{array}{lll}
\displaystyle \int \sin(ax)e^{bx} dx &=&\d...
... x)e^{b x} - \frac{a}{b}\int \cos(a x)e^{b x} dx
\end{array}\;.\end{displaymath}

Hence

\begin{displaymath}\int \cos(a x)e^{b x} dx = \frac{1}{b}\cos(a x)e^{b x} + \fra...
...in(a x)e^{b x} - \frac{a}{b}\int \cos(a x)e^{b x} dx\right] \;.\end{displaymath}

which implies

\begin{displaymath}\left(1 + \frac{a^2 }{b^2 }\right) \int \cos(a x)e^{b x} dx = \frac{1}{b}\cos(a x)e^{b x} + \frac{a}{b^2 }\sin(ax)e^{b x} \end{displaymath}

or

\begin{displaymath}\int \cos(ax)e^{bx} dx = \frac{b}{a^2 + b^2 } \cos(a x)e^{b x} + \frac{a}{a^2 + b^2 } \sin(a x)e^{b x} + C\;.\end{displaymath}

Detailed Answer.


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