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\large 
\underline
{MA2160 \hspace{5mm}Calculus II with Technology \hspace{50mm} Ch. 7-2}


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\Large
Goal
\large
\begin{enumerate}
\item Define integration by parts
\item Relate integration by parts to the product rule of differentiation
\item Solve problems using integration by parts

\end{enumerate}
What is integration by parts?  Integration by parts is defined by reverse engineering the product rule of differentiation:

$$ \frac{d}{dx}uv=uv'+u'v$$

where $u$ and $v$ are functions of $x$

If we take the integral of this equation, we get:

$$uv=\int\,uv'\,dx+\int\,u'v\,dx$$

rewriting:

$$\int\,uv'\,dx=uv-\int\,u'v\,dx$$

This is the formula for integration by parts.  This formula is used when we have an integral of the form:

$$\int\,f(x)g(x)\,dx$$

What is now required is to substitute $f(x)$ and $g(x)$ for $u$ and $v$.  The question now exists what principles guide the assignment to $u$ and $v$.  In general, use the following guidelines:

\begin{enumerate}
\item Choose as $u$ something that differentiates to 0 or something simpler than $u$
\item Choose as $v$ something that integrates indefinitely and stays the same complexity (if possible)
\end{enumerate}

As a simple example, find the following integral:

$$\int\,x \sin x\,dx$$

Letting $u=x$ and $v'=\sin x$, we can fill plug in to the formula:

$$\int\,x \sin x\,dx=-x\cos x- \int 1*-\cos x\, dx=-x\cos x + \sin x$$

The usefulness of this formula is that it can be applied multiple times to find a solution.

Problems to work on:  Examples 4, 5, 7; Exercises 18, 25, 43
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