Problems on Techniques of Integration

Use the Integration by Parts. Set

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

Then

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

So

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

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

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

Then

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

So

\begin{displaymath}\int \sin(x)e^x dx = \sin(x)e^x - \int \cos(x)e^x dx \;.\end{displaymath}

Hence

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

which implies

\begin{displaymath}2 \int \cos(x)e^x dx = \cos(x)e^x + \sin(x)e^x\end{displaymath}

or

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

Detailed Answer.


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