Integrals are puzzles!

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Two young mathematicians discuss how tricky integrals are puzzles.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Yo Riley, is it just me, or are integrals kind of fun?
Riley: I always feel accomplished when I finish one.
Devyn: I know! Also, even though antiderivatives are difficult, we can always check our work by taking the derivative.
Riley: So awesome!
Devyn: But something is bothering me. When we are doing substitution, we have to find $f$ and $u$ such that $\int f(u(x)) \cdot u'(x) \d x = \int f(u) \d u.$ How do we choose $f$ and $u$ ?
Riley: Well, never ever pick $u(x) = x$ , this doesn’t change anything!
Devyn: And never ever pick $u(x)$ to be the entire integrand, this doesn’t help either.
Riley: Somehow we must ‘‘see’’ one function ‘‘nested’’ inside of another.
Devyn: I’m not sure there’s an easy path to doing, this, I think it’s gonna take practice.

In the problems that follow, we will be using the substitution formula $\int f(u(x)) \cdot u'(x) \d x = \int f(u) \d u$ While you may use a slightly different method to compute your integrals, the skills developed by answering the problems below will help you in your quest to conquer calculus.

Consider $\int \sin ^5(3x) \cos (3x) \d x = \int f(u(x)) \cdot u'(x) \d x$ if $u(x) = 3x$ , and $\int f(u(x)) \cdot u'(x) \d x = \int f(u) \d u.$ what is $f(u)$ ? $f(u) = \answer { \frac {\sin ^5(u) \cos (u)}{3}}$

But we still have not yet integrated... Now we have $\int \frac {\sin ^5(u) \cos (u)}{3} \d u$

Consider $\int \sin ^5(3x) \cos (3x) \d x = \int f(u(x)) \cdot u'(x) \d x$ if $u(x) = \sin (3x)$ , then $du=3\cos (3x)\Rightarrow \frac {1}{3}du=\cos (3x)dx$ $\int \frac {1}{3} u^5 \d u=\frac {1}{3}\cdot \answer [given]{\frac {1}{6}\sin ^6(3x)}+C.$

In your own words, explain why Devyn and Riley claim we should never pick $u(x) = x$ or $u(x)$ to be the entire integrand.

The goal of substitution is to make the integral easier to do. Your choices for $f$ and $u$ should make things easier, not harder!

Unless the derivative of $u(x)$ is $1$ , choosing $u(x)$ to be the entire integrand means that you don’t have any part of the integrand left to be the derivative of $u$ . Choosing $u(x) = x$ means that $u'(x) = 1$ , meaning that you haven’t simplified the integral at all.