# 3.7: Integration Techniques

**At Grade**Created by: CK-12

## Integration by Substitution

The technique of substitution is the most important trick that is used for integration. The key to recognizing when substitution is best used is to look for the most complicated part of the function that involves the variable \begin{align*}x\end{align*}

\begin{align*}\int \frac{x^3}{\sqrt{1-x^4}}dx\end{align*}

The denominator is clearly the most complicated part so we focus on the portion \begin{align*}1-x^4\end{align*}. The derivative of this is \begin{align*}-4x^3\end{align*}, and this looks a lot like the numerator. So we try the substitution \begin{align*}u = 1-x^4\end{align*} to give \begin{align*}du=(-4x^3 )dx\end{align*} which means that:

\begin{align*}\frac{1}{-4} du = x^3 dx\end{align*}

So the integral reduces to:

\begin{align*}\frac{1}{-4} \int \frac{1}{\sqrt{u}} du\end{align*}

which is simple since \begin{align*}\int \frac{1}{\sqrt{u}} du = \int u^{\frac{1}{2}} du = \frac{3}{2} u^{\frac{3}{2}}\end{align*}.

## Integration by Parts

Integration by parts can be attempted with any integral whatsoever. The best way to try is simply to guess and check, since applying the rules methodically can be exhausting sometimes. The way to get good is to begin by practicing the product rule:

\begin{align*}(fg)' = f' g + fg'\end{align*}

So take any integral whatsoever, like:

\begin{align*}\int (x-8)^7 dx\end{align*}

You probably know two ways to do this integral, one is by expanding and the other is by a substitution. However, if neither of these occur to you might do something crazy like recognize that one of the terms from \begin{align*}\frac{d}{dx}(x(x-8)^7)\end{align*} is the integrand. That is:

\begin{align*}\frac{d}{dx} (x(x-8)^7 ) = (x-8)^7 + 7x(x-8)^6\end{align*}

So \begin{align*}x(x-8)^7\end{align*} almost works as an antiderivative, except for that pesky second term. Well so, all we have to do is find an antiderivative for this term and we’re done. This is the essence of integration by parts: Guess at an antiderivative that gives the integrand as one term from the product rule, and then go from there.

## Integration by Partial Fractions

Becoming efficient at decomposing fractions as partial fractions is a great skill to have for any scientist. Being able to decompose simple fractions mentally allows for a mental flexibility that is useful when looking at complicated formulas.

Suppose that the fraction \begin{align*}\frac{P}{Q}\end{align*} has a multiplicity of roots in the denominator so that there are some terms in the expansion like:

\begin{align*}\frac{A}{(x-a)} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}\end{align*}

we can of course combine these as:

\begin{align*}\frac{A(x-a)^2 + B (x-a) + C}{(x-a)^3}\end{align*}

and it turns out that the numerator, when expanded, is actually the Taylor Polynomial for \begin{align*}\frac{P}{(x-a)^3}\end{align*} of order \begin{align*}2\end{align*}. In general the object in the numerator of the expansion is always the Taylor Polynomial of order one less than the multiplicity.

## Trigonometric Integrals

Complicated trigonometric integrals appear in many areas of physics and engineering. One area where they are found very commonly is in experimental particle physics during calculations of *cross-sections.* Essentially the idea is to find the number of particles that will bounce into a certain direction given that they undergo a collision somewhere. Since the detectors are usually spherical, there are often a number of trigonometric factors that must be integrated over simultaneously.

Another important application for trigonometric integrals is in the area of frequency analysis which is used by sound engineers, physicists, and even by business theorists. In this theory a signal is broken into its individual *frequencies.* This involves a topic called Fourier analysis, which rests upon integration over many powers of sines and cosines.

## Trigonometric Substitutions

Trigonometric substitutions should be thought of as tricks that can be very useful for solving a variety of integrals. The only way to recognize when a trigonometric substitution should be applied is through experience. On the other hand, these tricks must be applied to a variety of integrals if they are to be solved.

Often the underlying cause for needing a trig substitution is that the function is best described in a different coordinate system. The only coordinates that have been discussed in this text are Cartestian or Rectangular coordinates. However, Polar Coordinates are also very useful in many contexts. These are described by the following diagram:

where instead of using the distances \begin{align*}x\end{align*} and \begin{align*}y (x,y)\end{align*} to describe a point’s location, one would use the point’s distance from the origin \begin{align*}r\end{align*} and the angle \begin{align*}\theta\end{align*} formed with the \begin{align*}x-\end{align*}axis. As an example of how useful these coordinates can be, consider a circle of radius \begin{align*}R\end{align*}. In Cartesian Coordinates the circle can be described as the set of points \begin{align*}(x,y)\end{align*} satisfying \begin{align*}x^2+y^2=R^2\end{align*}, whereas in polar coordinates the circle is just \begin{align*}r=R\end{align*}. It is not a coincidence that the integral of area for a circle is most efficiently done by using a trigonometric substitution.

## Improper Integrals

Integrals with infinite limits or that pass over points of discontinuity are extremely important in physics as well as engineering. It is actually kind of miraculous that the area under a curve like the one below is not infinite:

This area is not infinite even though the tail of the graph goes off to the right all the way out the infinity and is never zero. Similarly, a graph can go off to infinity in the vertical direction but still have an area that is not infinite. This is part of what is nice about having numbers. The picture looks like the area is infinity, but the math gives us a different answer that we can use.

## Ordinary Differential Equations

How does calculus appear in actual physics or engineering applications? The answer is through differential equations. Basically a theory tells us what the differential equation should look like, and then to find out what will happen we need to either solve the equation or else numerically approximate a solution based on the starting conditions. Often the result depends very delicately upon the starting conditions, and when the result is drastically different for different starting conditions we have chaos.

Most of the time scientists will be able to look at a differential equation and understand some basic ideas about how the function behaves. For example, suppose we think abstractly about the simple differential equation:

\begin{align*}\frac{dy}{dx}=10\end{align*}

This tells us that no matter where \begin{align*}y\end{align*} starts, it will be increasing by \begin{align*}10\end{align*} for every step in \begin{align*}x\end{align*}. So we can tell it will be a line sloped upwards, although this line could be anywhere in the plane:

The different possible solutions (each of the lines above) simply represent different possible values for the constant.

Now consider the slightly more complicated differential equation:

\begin{align*}\frac{dy}{dx}=x\end{align*}

Here you can imagine that the slope is increasing linearly with \begin{align*}x\end{align*}, and so we must have a parabola. Again though, it could be located anywhere so we’d get a series of parabolas sitting on top of one-another.

Ordinary differential equations (ODEs) contain a function \begin{align*}y\end{align*} of one real variable \begin{align*}x\end{align*}. However, the right hand side of:

\begin{align*}\frac{dy}{dx}=F(x,y)\end{align*}

is a general function \begin{align*}F\end{align*} of **two** variables \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. So the most general ODE requires the ability to analyze functions of more than one variable, in order to completely understand the behavior of the equation. This, of course, is only the case when \begin{align*}F\end{align*} is not separable and no other simple trick applies to obtain an exact solution.

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