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 $\int f(g(x)) \cdot g'(x) \d x = \int f(g) \d g.$ 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 $\int f(g(x)) \cdot g'(x) \d x = \int f(g) \d g$ 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.

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.