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 $g$ such that $\int f(g(x)) \cdot g'(x) dx = \int f(g) dg.$ How do we choose $f$ and $g$ ?
Riley: Well, never ever pick $g(x) = x$ , this doesn’t change anything!
Devyn: And never ever pick $g(x)$ 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 $\int f(g(x)) \cdot g'(x) dx = \int f(g) dg$ 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 $\int \sin ^5(3x) \cos (3x) dx = \int f(g(x)) \cdot g'(x) dx$ if $g(x) = 3x$ , and $\int f(g(x)) \cdot g'(x) dx = \int f(g) dg.$ what is $f(g)$ ? $f(g) = \answer { \frac {\sin ^5(g) \cos (g)}{3}}$

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

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

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

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