# Antidifferentiation

Since a lot of classes start integration with antidifferentiation, I’ll discuss that first. If you decided to go with one of the other plans mentioned in my last post, then file this away for later.
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The key to finding antiderivatives is pattern recognition.

The simplest integrals are those that follow directly from derivatives such as. Students just need to recognize that the cosine is the derivative of the sine so $\int{\cos \left( x \right)dx}=\sin \left( x \right)+C$. We wish they were all that simple.

Something like $\int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx}$ is more complicated. Is the integrand the derivative of $\tfrac{1}{2}{{\tan }^{2}}\left( x \right)$  or of $\tfrac{1}{2}{{\sec }^{2}}\left( x \right)$? The answer is yes.

The method called u-substitution helps in identifying the pattern of the integrand.  To use this method identify some part of the integrand as a function which you call u and then calculate du and hope that the du is the rest of the integrand.

For $\int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx}$ you can try u = tan(x) and the du = sec2(x)dx and after making the substitutions:

$\int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx}=\int{udu}=\tfrac{1}{2}{{u}^{2}}=\tfrac{1}{2}{{\tan }^{2}}\left( x \right)+C$.

On the other hand we could try u = sec(x) so that du = tan(x)sec(x)dx and

$\int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx}=\int{\sec (x)\left( \tan (x)\sec (x)dx \right)}$

$=\int{udu}=\tfrac{1}{2}{{u}^{2}}=\tfrac{1}{2}{{\sec }^{2}}\left( x \right)+C*$

(The + C* here not the same as the + C in the expression above: C* = C – ½.)

Either way students need to recognize that part of the integrand is the chain rule contribution to the derivative of the other part of the integrand. The usual three steps to acquire these pattern recognition skills are practice, practice, practice.

Another detail of u-substitution is handling missing constants. For example, $\int{\cos (2x)dx}$ .

Make the substitution u = 2x and then calculate du = 2dx so $dx=\tfrac{1}{2}du$. Then making these substitutions

$\int{\cos (2x)dx}=\int{\cos \left( u \right)\left( \tfrac{1}{2}du \right)=}\tfrac{1}{2}\int{\cos \left( u \right)du=\tfrac{1}{2}\sin \left( u \right)}$.

Then back-substituting: $\int{\cos (2x)dx}=\tfrac{1}{2}\sin \left( 2x \right)+C$.

For simple u-substitutions like this I find it easier to think, and not write, u = 2x so du = 2dx and then write this multiplying by ½ outside the integral sign to account for the extra factor of 2.

$\int{\cos \left( 2x \right)dx=\tfrac{1}{2}\int{\cos \left( 2x \right)\left( 2dx \right)}}=\tfrac{1}{2}\sin \left( 2x \right)+C$.

This idea only works with missing constants, since only constants can be moved in and out of integrands.

I explain the need for the constant to students by saying that the 2 appears (as if by magic) from the Chain Rule when you differentiate and therefore it must be present to disappear back into the Chain Rule when you antidifferentiate.

Point out to students that integration is very different than differentiation. Differentiation is pretty straightforward; you know what to do when you need to find a derivative. Antidifferentiation is more complicated since recognizing the form or pattern is necessary. The simpler looking integral $\int{\ln \left( x \right)}dx$ is really more difficult than $\int{\tfrac{\ln \left( x \right)}{x}dx}$.

Finally, if you are teaching antiderivatives before beginning integration, when you get to definite integrals, you will have to remember to show students how to handle the limits of integration by using the same substitution.

For Advanced Placement AB calculus courses these (integrals that follow directly from derivatives and u-substitutions) are the only methods of integration tested. BC students also need to know integration by parts and partial fraction decomposition. I will discuss these later.