Problems on Techniques of Integration

The trigonometric functions are nice functions since their derivatives and antiderivatives are easy to get. So for the integration by parts, these functions have the same behavior whether we differentiate them or take their antiderivatives. Therefore the focus should be on the other function involved in the integration. Since the derivative or antiderivative of $e^x$ is even easier, so in this case we may either differentiate or integrate, it will not make any major difference.

Set

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

Then

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

Since

\begin{displaymath}\int u dv = u v - \int v du\;,\end{displaymath}

we get

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

In order to integrate the function $\cos(x)e^x$, we will need to do another integration by parts. This is very common that an integration by parts may lead to another one or may be more integration by parts. Set

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

Then

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

Since

\begin{displaymath}\int u dv = u v - \int v du\;,\end{displaymath}

we get

\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}

Hence

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

which implies

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

or

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

It may be surprising that the constant $C$ showed up at the end and not before. In fact, it has to do with the formula

\begin{displaymath}\int u dv = u v - \int v du\;.\end{displaymath}

Indeed, when obtaining it, we had to take the antiderivative of $(uv)'$ which is not $uv$ but $uv + C$. In any case, one should not forget the constant $C$ when dealing with antiderivatives or indefinite integrals.


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