Integrals are puzzles!

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\descriptionlabel }[1]{\hspace {\labelsep }\textbf {#1:}}$

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.

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

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