The idea of substitution

$\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 }$

We learn a new technique, called substitution, to help us solve problems involving integration.

Computing antiderivatives is not as easy as computing derivatives. One issue is that the chain rule can be difficult to “undo.” We have a general method called “integration by substitution” that will somewhat help with this difficulty.

If the functions are differentiable, we can apply the chain rule and obtain $\ddx f(g(x)) = f'(g(x))g'(x)$ If the derivatives are continuous, we can use this equality to evaluate a definite integral. Namely,

$\begin{align*} \int_a^b f'(g(x))g'(x) \d x &= \eval{f(g(x))}_a^b \\ &= f(g(b)) - f(g(a)). \\ \end{align*}$

On the other hand, it is also true that

$\begin{align*} \int_{g(a)}^{g(b)} f'(u)\d u &= \eval{f(u)}_{g(a)}^{g(b)}\\ &= f(g(b)) - f(g(a)). \\ \end{align*}$

Since the right hand sides of these equalities are equal, the left hand sides must be equal, too. So, it follows that

$\int _a^b f'(g(x))g'(x) \d x = \int _{g(a)}^{g(b)} f'(u)\d u .$

This simple observation leads to the following theorem.

Integral Substitution Formula Let $g'$ be the derivative of a function $g$ and let $f'$ be the derivative of a function $f$ . If $g'$ is continuous on the interval $[a,b]$ and $f'$ is continuous on the interval $[g(a),g(b)]$ , then $\int _a^b f'(g(x)) g'(x) \d x =\int _{g(a)}^{g(b)} f'(u) \d u.$

The integral on the right appears to be much simpler than the original integral. So, when evaluating a difficult integral, we try to apply this theorem and replace the original integral with a simpler one. We can do this as long as the conditions of the theorem are satisfied.

We will apply the Integral Substitution Formula (ISF) to the following example.

Compute: $\int _0^2 x e^{x^2}\d x$

Why do we think that the ISF can be useful in this example?

Does this integral have the structure $\int _a^b f'(g(x)) g'(x) \d x ,$ so that we can apply the ISF?

We set $g(x) =x^2$ , so $g'(x)dx =2xdx$ , and note that

$\int _0^2 x e^{x^2}\d x =\frac {1}{2}\int _0^{2} e^{\underbrace {x^2}}\underbrace {2x\d x}=\frac {1}{2}\int _0^{2} e^{g(x)}g'(x)\d x$ So we wrote the integral in a form suitable for application of the ISF. Therefore, we can substitute $u$ for $g(x)$ and $\d u$ for $g'(x)\d x$ and write

$\int _0^2 x e^{x^2}\d x=\frac {1}{2}\int _0^{2} e^{g(x)}g'(x)\d x=\frac {1}{2}\int _{g(0)}^{g(2)} e^{u}\d u$ Now, compute the endpoints:

$\begin{align*} g(0) &= 0^2 = 0 \\ g(2) &=2^2 = 4. \end{align*}$

We can easily evaluate the last integral $\frac {1}{2}\int _{g(0)}^{g(2)} e^{u}\d u= \frac {1}{2}\int _{0}^{4} e^{u}\d u=\frac {1}{2}\eval {e^u}_0^4=\frac {1}{2}\left (\answer [given]{e^4}-e^0\right )=\frac {\answer [given]{e^4}-1}{2}$ In summary,

$\int _0^2 x e^{x^2}\d x=\frac {1}{2}\int _{0}^{4} e^{u}\d u$ The figure below illustrates both integrals.

The two shaded areas in the figure, $A= \int _0^2 x e^{x^2}\d x$ and $B=\frac {1}{2}\int _{0}^{4} e^{u}\d u$ , appear to be equal, and this confirms what was established by computation before.

Therefore, whenever we are faced with a problem of evaluating a difficult integral, we try to replace it with a simpler one, as long as both integrals represent the same net area.

We can solve a problem like this in a slightly different way. Let’s do the same example again, this time we will think in terms of differentials.

Compute: $\int _0^2 x e^{x^2}\d x$

Here we will set $u=g(x) = x^2.$ Then $\d u =g'(x)\d x=2x\d x.$ Here, we are thinking in terms of differentials. We can solve for $\d x$ to get $\d x = \frac {\d u}{2x}.$ Using this equality, we can write $\begin{align*} \int_0^2 x e^{x^2}\d x &= \int_{g(0)}^{g(2)} x e^{u} \frac{\d u}{2x}\\ &= \int_{0}^{4} \frac{e^u}{2}\d u . \end{align*}$

At this point, we can continue as we did before and write $\int _{0}^{4} \frac {e^u}{2}\d u= \frac {\answer [given]{e^4} -e^0}{2}.$

Finally, sometimes we simply want to deal with the antiderivative on its own. We will repeat the example one more time in oder to demonstrate this approach.

Compute: $\int _0^2 x e^{x^2}\d x$

We can apply the Second Fundamental Theorem of Calculus and, in order to do that, we first need to compute an indefinite integral. $\int x e^{x^2} \d x.$

As before, we set $u=g(x)=x^2$ , and compute $\d u = 2x \d x$ , thinking in terms of differentials. Now we see that $\int x e^{x^2} \d x = \int x e^{u} \frac {\d u}{2x} = \int \frac {e^{u}}{2}\d u .$ Hence $\int x e^{x^2} \d x = \frac {e^{u}}{2}+C .$ This does not look right, since we need to find an antiderivative of the function $x e^{x^2},$ and our result is a function of $u$ . We can easily fix this, by simply substituting $u$ with $g(x)=x^2$ . Therefore, $\int xe^{x^2}\d x =\frac {e^{u}}{2}+C= \frac {e^{x^2}}{2}+C .$ Finally, we can apply the SFTOC

$\begin{align*} \int_0^2 x e^{x^2}\d x &=\eval{\frac{e^{x^2}}{2}}_0^2\\ &= \frac{\answer[given]{e^4} -e^0}{2}. \end{align*}$

More examples

With some experience, it is (usually) not too hard to see what to substitute as $u$ . We will work through the following examples in the same way that we did for Example main example.

Compute: $\int x^4(x^5+1)^{9} \d x$

Here we set $u = \answer [given]{x^5+1}$ , so $\d u = \answer [given]{5x^4} \d x$ . Then it follows that $\d x=\frac {\d u}{5x^4}$ . Therefore $\begin{align*} \int x^4(x^5+1)^{9} \d x &= \int x^4 (u)^{9} \frac{\d u}{5x^4} \\ &= \frac{1}{5} \int u^{9} \d u\\ &=\frac{u^{10}}{ \answer[given]{50}}. \end{align*}$

Notice that this example is an indefinite integral and not a definite integral, meaning that there are no limits of integration. So we do not need to worry about changing the endpoints of the integral. However, we do need to back-substitute into our answer, so that our final answer is a function of $x$ . Recalling that $u= x^5+1$ , we have our final answer $\int x^4(x^5+1)^{9} \d x= \frac {(x^5+1)^{10}}{\answer [given]{50}}+C.$ Reminder: you can always verify your result by differentiating.

If substitution works to solve an integral (and that is not always the case!), a common trick to find what to substitute for is to locate the “ugly” part of the function being integrated. We then substitute for the “inside” of this ugly part. While this technique is certainly not rigorous, it can prove to be very helpful. This is especially true for students new to the technique of substitution. The next two problems are really good examples of this philosophy.

Compute: $\int _{-1}^0 12x^3 \cos ({x^4}) \d x$

The “ugly” part of the function being integrated is $\cos ({x^4})$ . The “inside” of this term is then $x^4$ . So a good possibility is to try $u =g(x)= x^4.$ Then $\d u = (4x^3) \d x \qquad \Rightarrow \qquad \d x = \answer [given]{\frac {1}{4x^3}} \d u$ and so $\begin{align*} \int_{-1}^0 12x^3 \cos({x^4}) \d x &= \int_{g(-1)}^{g(0)} 12 x^3 \cos({u}) \answer[given]{\frac{1}{4x^3}} \d u \\ &= \int_{\answer[given]{1}}^{\answer[given]{0}} 3 \cos({u}) \d u \\ &= \eval{3\sin({u})}_{\answer[given]{1}}^{\answer[given]{0}} \\ &= \answer[given]{3(0-\sin({1}))}. \end{align*}$

Compute: $\int _{1}^{e^{\frac {\pi }{4}}} \frac {\cos (\ln x)}{x} \d x$

Here the “ugly” part here is $\cos (\ln x)$ . So we substitute for the inside: $u=g(x)=\answer [given]{\ln x}.$ Then $\d u = \answer [given]{\frac {1}{x}} \d x \qquad \Rightarrow \qquad \d x = \answer [given]{x} \d u.$ Notice that $\begin{align*} g(1) &= \ln (1) = \answer[given]{0} \\ g \left( e^{\frac{\pi}{4}} \right) &= \ln \left( e^{\frac{\pi}{4}} \right) = \answer[given]{\frac{\pi}{4}}. \end{align*}$

Then we substitute back into the original integral and solve:

$\begin{align*} \int_{1}^{e^{\frac{\pi}{4}}} \frac{\cos(\ln x)}{x} \d x &= \int_0^{\frac{\pi}{4}} \frac{\cos(u)}{x} x \d u \\ &= \int_0^{\frac{\pi}{4}} \cos(u) \d u \\ &= \eval{\answer[given]{\sin(u)}}_{0}^{\frac{\pi}{4}} \\ &= \frac{\sqrt{2}}{\answer[given]{2}} - \answer[given]{0} = \frac{\sqrt{2}}{\answer[given]{2}}. \end{align*}$

To summarize, if we suspect that a given function is the derivative of another via the chain rule, we introduce a new variable $u=g(x)$ , where $g$ is a likely candidate for the inner function. We rewrite the integral entirely in terms of $u$ , with no $x$ remaining in the expression. If we can integrate this new function of $u$ , then the antiderivative of the original function is obtained by replacing $u$ by $g(x)$ .