Working with substitution

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \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 {\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 {\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 {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\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 {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We explore more difficult problems involving substitution.

We begin by restating the substitution formula.

Integral Substitution Formula If $g$ is differentiable on the interval $[a,b]$ and $f$ is differentiable on the interval $[g(a),g(b)]$ , then $\int _a^b f'(g(x)) g'(x) \d x =\int _{g(a)}^{g(b)} f'(g) \d g.$

We spend pretty much this entire section working out examples.

Compute: $\int _{2}^{3} \frac {1}{x\ln (x)} \d x$

Let $g =\answer [given]{\ln (x)},$ computing $\d g$ , we find $\d g =\answer [given]{\frac {1}{x}}\d x$ and solving for $\d x$ we find $\d x = \answer [given]{x} \d g.$ Now $\begin{align*} \int_{2}^{3} \frac{1}{x\ln(x)} \d x &= \int_{g(2)}^{g(3)} \frac{1}{x \cdot g} x \d g \\ &= \int_{\answer[given]{\ln(2)}}^{\answer[given]{\ln(3)}} \answer[given]{\frac{1}{g}} \d g\\ &= \eval{\answer[given]{\ln(g)}}_{\answer[given]{\ln(2)}}^{\answer[given]{\ln(3)}}\\ & = \ln(\ln(3)) - \ln(\ln(2)). \end{align*}$

The next example requires a new technique.

Compute: $\int x^3\sqrt {1-x^2}\d x$

Here it is not apparent that the chain rule is involved. However, if it was involved, perhaps a good guess for $g$ would be $g = \answer [given]{1-x^2}$ and then $\begin{align*} \d g &= \answer[given]{-2x} \d x, \\ \d x &= \answer[given]{\frac{-1}{2x}} \d g. \end{align*}$ Now we consider the integral we are trying to compute $\int x^3\sqrt {1-x^2}\d x$ and we substitute using our work above. Write with me $\begin{align*} \int x^3\sqrt{1-x^2}\d x &= \int x^3 \answer[given]{\sqrt{g}} \left( \answer[given]{\frac{-1}{2x}} \right) \d g \\ &= \int \frac{-x^2 \answer[given]{\sqrt{g}}}{2}\d g. \end{align*}$ However, we cannot continue until each $x$ is replaced. We know that $\begin{align*} g &= 1-x^2 \\ \Rightarrow \qquad g -1 &= -x^2\\ \Rightarrow \qquad \answer[given]{1- g} &= x^2 \end{align*}$ so now we may replace $x^2$ $\int x^3\sqrt {1-x^2}\d x = \int -\frac {\answer [given]{(1-g)} \sqrt {g}}{2}\d g.$ At this point, we are close to being done. Write $\begin{align*} \int -\frac{(1-g)\sqrt{g}}{2}\d g &= \int \left(\frac{g\sqrt{g}}{2} - \frac{\sqrt{g}}{2}\right) \d g \\ &= \int \frac{\answer[given]{g^{3/2}}}{2} \d g - \int \frac{\sqrt{g}}{2} \d g \\ &= \answer[given]{\frac{g^{5/2}}{5}} - \answer[given]{\frac{g^{3/2}}{3}}. \end{align*}$ Now recall that $g = 1-x^2$ . Hence our final answer is $\int x^3\sqrt {1-x^2}\d x = \answer [given]{\frac {(1-x^2)^{5/2}}{5}} - \frac {(1-x^2)^{3/2}}{3}+C.$

Sometimes it is not obvious how a fraction could have been obtained using the chain rule. A common trick though is to substitute for the denominator of a fraction. Like all tricks, this technique does not always work. Regardless the next two examples present how this technique can be used.

Compute: $\int \frac {\sec (y) \tan (y) + \sec ^2(y)}{\sec (y) + \tan (y)} \d y$

We substitute $g = \sec (y) + \tan (y)$ and we immediately see that $\begin{align*} \d g &= \answer[given]{(\sec(y) \tan(y) + \sec^2(y))} \d y,\\ \d y &= \frac{1}{\answer[given]{\sec(y) \tan(y) + \sec^2(y)}} \d g. \end{align*}$ But this cancels perfectly with the numerator! So we have that $\begin{align*} \int \frac{\sec(y) \tan(y) + \sec^2(y)}{\sec(y) + \tan(y)} \d y &= \int \frac{1}{g} \d g \\ &= \ln|g| + C \\ &= \ln| \sec(y) + \tan(y) | + C. \end{align*}$

Notice that $\frac {\sec (x) \tan (x) + \sec ^2(x)}{\sec (x) + \tan (x)} = \frac {\sec (x) (\tan (x) + \sec (x))}{\sec (x) + \tan (x)} = \sec (x)$ when $\sec (x) \neq - \tan (x)$ . So in a very contrived way, we have just proved

$\int \sec (x) \d x = \ln |\sec (x) + \tan (x)| + C.$

Notice the variable in this next example.

Compute: $\int \frac {g}{1-g^2} \d g$

We want to substitute for $1-g^2$ . But the variable “ $g$ ” has already been used…OH NO! Never fear! We can substitute with whatever variable that we want. In particular, let us use “ $h$ ” for this problem. So we let $h = 1 - g^2$ and then $\begin{align*} \d h &= \answer[given]{-2g} \d g,\\ \d g &= \answer[given]{\frac{-1}{2g}} \d h. \end{align*}$ Thus $\begin{align*} \int \frac{g}{1-g^2} \d g &= \int \frac{g}{h} \left( \answer[given]{\frac{-1}{2g}} \right) \d h \\ &= - \frac{1}{2} \int \frac{1}{h} \d h \\ &= - \frac{1}{2} \ln|h| + C \\ &= - \frac{1}{2} \ln|1-g^2| + C. \end{align*}$

Compute: $\int \tan (x) \d x$

We begin by writing $\begin{align*} \tan(x) &= \frac{\sin(x)}{\cos(x)} \\ &= \frac{\sin(x)}{\cos(x)} \cdot \frac{\cos(x)}{\cos(x)} \\ &= \frac{\sin(x) \cos(x)}{1 - \sin^2(x)}. \end{align*}$ We then make the substitution $g = \sin (x)$ and so $\begin{align*} \d g &= \answer[given]{\cos(x)} \d x,\\ \d x &= \frac{1}{\answer[given]{\cos(x)}} \d g. \end{align*}$ Then $\begin{align*} \int \tan(x) \d x &= \int \frac{\sin(x) \cos(x)}{1 - \sin^2(x)} \d x \\ &= \int \frac{g \cos(x)}{1 - g^2} \cdot \frac{1}{\answer[given]{\cos(x)}} \d g \\ &= \int \frac{g}{1-g^2} \d g. \end{align*}$ But this is the same problem as Example ??! And so we know that $\begin{align*} \int \tan(x) \d x &= - \frac{1}{2} \ln|1-g^2| + C \\ &= - \frac{1}{2} \ln|1-\sin^2 (x)| + C \\ &= - \frac{1}{2} \ln|\cos^2(x)| + C \\ &= \ln|\cos^2(x)|^{- \frac{1}{2}} + C \\ &= \ln|\sec(x)| + C. \end{align*}$

We have just proved

$\int \tan (x) \d x = \ln |\sec (x)| + C.$

Note that in Example ??, we could have instead made the substitution $g = 1-\sin ^2(x).$ This would have gotten us to the answer quicker and without using Example ??. You are encouraged to work this out on your own right now!

We end this section with two more difficult examples.

Compute: $\int \frac {e^{2x}}{1 - e^{2x}} \d x$

Maybe the biggest key to solving this problem is to recall that $e^{2x} = (\answer [given]{e^x})^2.$ So we can rewrite the problem $\int \frac {e^{2x}}{1 - e^{2x}} \d x = \int \frac {(\answer [given]{e^x})^2}{1 - (\answer [given]{e^x})^2} \d x.$ Now, if we make the substitution $g = e^x$ , we have that $\begin{align*} \d g &= \answer[given]{e^x} \d x,\\ \d x &= \frac{1}{\answer[given]{e^x}} \d g, \end{align*}$ and $\begin{align*} \int \frac{e^{2x}}{1 - e^{2x}} \d x &= \int \frac{(e^x)^2}{1 - g^2} \cdot \frac{1}{\answer[given]{e^x}} \d g \\ &= \int \frac{e^x}{1-g^2} \d g \\ &= \int \frac{g}{1-g^2} \d g. \end{align*}$ But now we are back to Example ??, and so we know that $\begin{align*} \int \frac{e^{2x}}{1 - e^{2x}} \d x &= - \frac{1}{2} \ln|1-g^2| + C \\ &= - \frac{1}{2} \ln|1 - e^{2x}| + C. \end{align*}$

Again, in the previous example we could have instead made the substitution $g = 1 - e^{2x}$ and avoided using Example ??. In general, any time that you make two successive substitutions in a problem, you could have instead just made one substitution. This one substitution is the composition of the two original substitutions. But sometimes it may not be obvious to make one clever substitution, and so two substitutions makes more sense. The next example helps to demonstrate this.

Compute: $\int _0^{16} \sqrt {4 - \sqrt {x}} \d x$

While it is not obvious at all, let us try the substitution $g = \sqrt {x}.$ Then $\begin{align*} \d g &= \answer[given]{\frac{1}{2 \sqrt{x}}} \d x,\\ \d x &= 2 \sqrt{x} \d g = 2g \d g, \end{align*}$ and so $\begin{align*} \int_0^{16} \sqrt{4 - \sqrt{x}} \d x &= \int_{g(0)}^{g(16)} \answer[given]{\sqrt{4-g} \cdot 2g} \d g \\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} \answer[given]{2g \sqrt{4-g}} \d g. \end{align*}$ From here we now make the second (and more obvious) substitution $h = 4-g.$ Then $g = 4-h$ , and $\begin{align*} \d h &= - \d g,\\ \d g &= - \d h. \end{align*}$ So $\begin{align*} \int_0^{16} \sqrt{4 - \sqrt{x}} \d x &= \int_0^4 2g \sqrt{4-g} \d g \\ &= \int_{h(0)}^{h(4)} 2 (4-h) \sqrt{h} (-1) \d h \\ &= - \int_{\answer[given]{4}}^{\answer[given]{0}} ( 8h^{\frac{1}{2}} - 2h^{\frac{3}{2}} ) \d h \\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} ( 8h^{\frac{1}{2}} - 2h^{\frac{3}{2}} ) \d h \\ &= \eval{ \answer[given]{8 \left( \frac{2}{3} \right) h^{\frac{3}{2}} - 2 \left( \frac{2}{5} \right) h^{\frac{5}{2}}} }_{0}^{4} \\ &= \left( \frac{16}{3} (4)^{\frac{3}{2}} - \frac{4}{5} (4)^{\frac{5}{2}} \right) - \left( 0 - 0 \right) \\ &= \frac{128}{3} - \frac{128}{5} \\ &= \frac{256}{15}. \end{align*}$