Substitution Rule for Integration
An important integration rule.
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Study Substitution Rule for Integration
We now collect some rules for the integration of more complicated functions. Unfortunately, integration is not as straightforward as differentiation and one often has to have an “inspired guess” to find out the antiderivative.
Integration by Substitution
Theorem 1 (Integration by Substitution). Let \(I\) be an Interval, \(f:I\to\mathbb{R}\) be continuous and \(\phi:[a,b]\to I\) be continuously differentiable. Then \[\int_a^bf(\phi(t))\phi'(t)\,dt=\int_{\phi(a)}^{\phi(b)}f(x)\, dx.\]
Proof: Let \(F: I\to\mathbb{R}\) be an antiderivative of \(f\). Then, according to the chain rule, the function \(F\circ\phi:[a,b]\to\mathbb{R}\) is differentiable with \[(F\circ\phi)'(t)=F'(\phi(t))\phi'(t)=f(\phi(t))\phi'(t).\] Therefore, \[\int_a^bf(\phi(t))\phi'(t)\,dt=\int_a^b(F\circ\phi)'(t)dt=\left.(F\circ\phi)(t)\right|_{t=a}^{t=b}=\left.F(x)\right|_{x=\phi(a)}^{x=\phi(b)}=\int_{\phi(a)}^{\phi(b)}f(x)\, dx.\] \(\Box\)
As a direct conclusion of this results, we can formulate the following:
Theorem 2 (Integration by Substitution II). Let \(I\) be an interval, \(g:I\to \mathbb{R}\) be continuously differentiable and injective with inverse function \(g^{-1}:g(I)\to \mathbb{R}\). Let \(f:J\to\mathbb{R}\) with \(J\subset g(I)\). Then \[\int_{a}^{b}f(x)\, dx=\int_{g^{-1}(a)}^{g^{-1}(b)}f(g(t))g'(t)\,dt.\]
Example 3. We can use the substitution rule to determine the area of an ellipse. The equation of an ellipse is given by \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\] This leads to \[y=\pm b\sqrt{1-\frac{x^2}{a^2}},\qquad x\in[-{a},a].\] As a consequence, the area of an ellipse is given by \[A=2\int_{-a}^ab\sqrt{1-\frac{x^2}{a^2}}\, dx=2b\int_{-a}^a\sqrt{1-\frac{x^2}{a^2}}\, dx\] Now we set \(g(t)=a\sin(t)\) and \(f(x)=\sqrt{1-\frac{x^2}{a^2}}\). According to the substitution rule, we now have \[\begin{aligned} \int_{-a}^a\sqrt{1-\frac{x^2}{a^2}}\, dx&=\int_{\arcsin\left(-\frac{a}a\right)}^{\arcsin\left(\frac{a}a\right)}\sqrt{1-\frac{a^2\sin^2(t)}{a^2}}(a\sin)'(t)\,dt\\ &=\int_{-\frac\pi2}^{\frac\pi2}\sqrt{1-\sin^2(t)}a\cos(t)\,dt=a\int_{-\frac\pi2}^{\frac\pi2}\cos^2(t)\,dt. \end{aligned}\] With \[\cos^2(t)=\frac14(\exp(it)+\exp(-it))^2=\frac14(\exp(2it)+2+\exp(-2it))=\frac12\cos(2t)+\frac12,\] we obtain \[\begin{aligned} A=&2\int_{-a}^ab\sqrt{1-\frac{x^2}{a^2}}\, dx =2ab\int_{-\frac\pi2}^{\frac\pi2}\cos^2(t)dt\\=&2ab\int_{-\frac\pi2}^{\frac\pi2}\frac12\cos(2t)+\frac12dt =ab\int_{-\frac\pi2}^{\frac\pi2}\cos(2t)+1dt\\ =&ab\cdot\left(\left.\frac{1}2\sin(2t)+t\right|_{t=-\frac\pi2}^{t=\frac\pi2}\right) =ab\cdot\left(\frac12\sin(\pi)-\frac12\sin\left(-\pi\right)+\frac\pi2+\frac\pi2\right)=\pi ab. \end{aligned}\]
Note that integration by substitution can also be applied by using the following formalism for determining \(\int_a^b f(x)\, dx\): Consider the substitution \(x=g(t)\) \(\Rightarrow\) \(g'(t)=\frac{d}{dt}g(t)=\frac{dx}{dt}\) and “a formal multiplication with \(dt\) yields \(dx=g'(t)dt\). For a formal determination of the integration bounds, we consider the equations \(a=g(t_l)\), \(b=g(t_u)\) and thus \(t_l=g^{-1}(a)\), \(t_u=g^{-1}(b)\). Integration by substitution can then be formally done by \[\underbrace{\int_a^b}_{\int_{g^{-1}(a)}^{g^{-1}(b)}} f(\underbrace{x}_{=g(t)})\underbrace{dx}_{=g'(t)dt}=\int_{g^{-1}(a)}^{g^{-1}(b)}f(g(t))g'(t)dt.\]
Example 4.
For \(a,b\in\mathbb{R}\), determine \[\int_a^bx^2\sin(x^3)dx.\] Consider the “new variable” \(t=x^3\). Then \(x=\sqrt[3]{t}=t^{1/3}\) and \(\frac{dx}{dt}=\frac13t^{-2/3}\) and thus \(dx=\frac13t^{-2/3}dt\). The integration bounds are given by \(t_l=a^3\) and \(t_u=b^3\) and thus \[\int_{a^3}^{b^3}t^{2/3}\sin(t)\frac13t^{-2/3}dt=\frac13\int_{a^3}^{b^3}\sin(t)dt=-\frac13\left.\cos(t)\right|_{t=a^3}^{t=b^3}=-\frac13\left.\cos(x^3)\right|_{x=a}^{x=b}.\]
For \(a\geq0\), \(b\geq0\), determine \[\int_a^b\exp(\sqrt{x})dx.\] Consider the substitution \(x=t^2\). Then \(\frac{dx}{dt}=2t\) and thus \(dx=2tdt\). For the integration bounds, consider \(a=t_l^2\) and \(b=t_u^2\) which yields \(t_l=\sqrt{a}\), \(t_u=\sqrt{b}\). We now get \[\begin{aligned} \int_a^b\exp(\sqrt{x})dx=&\int_{\sqrt{a}}^{\sqrt{b}}\exp(t)2tdt\\=&2\int_{\sqrt{a}}^{\sqrt{b}}t\exp(t)dt\\ =&\left.2\exp(t)(t-1)\right|_{t=\sqrt{a}}^{t=\sqrt{b}} =\left.2\exp(\sqrt{x})(\sqrt{x}-1)\right|_{x={a}}^{x={b}}. \end{aligned}\]
For \(a,b\in[-1,\infty)\), determine \[\int_a^b\frac{x^2+1}{\sqrt{x+1}}dx.\] Consider the “new variable” \(t=\sqrt{x+1}\). Then \(x=t^2-1\) and \(\frac{dx}{dt}=2t\Rightarrow dx=2tdt\) and \[\begin{aligned} \int_a^b\frac{x^2+1}{\sqrt{x+1}}dx&=\int_{\sqrt{a+1}}^{\sqrt{b+1}}\frac{(t^2-1)^2+1}{t}2tdt\\ &=\int_{\sqrt{a+1}}^{\sqrt{b+1}}2t^4-4t^2+4dt\\ &=\left.\frac25t^5-\frac43t^3+4t\right|_{t=\sqrt{a+1}}^{t=\sqrt{b+1}}\\ &=\left.\frac25\sqrt{x+1}^5-\frac43\sqrt{x+1}^3+4\sqrt{x+1}\right|_{x=a}^{x=b}. \end{aligned}\]
By using the substitution rule, we can also integrate expressions of type \(\frac{g'(x)}{g(x)}\).
For a differentiable function \(g:[a,b]\to\mathbb{R}\) with \(g(x)\neq0\) for all \(x\in[a,b]\) holds \[\int_a^b\frac{g'(x)}{g(x)}dx=\left.\log(|g(x)|)\right|_{x=a}^{x=b}.\]
Proof: For \(f(y)=\frac1y\), the above integral is of type \[\int_a^bf(g(x))g'(x)dx\] and thus, the result follows by the substitution rule.\(\Box\)
Example 5.
For \(a,b\in\left[-\frac\pi2,\frac\pi2\right]\) holds \[\begin{aligned} \int_a^b\tan(x)dx&=\int_a^b\frac{\sin(x)}{\cos(x)}dx=\int_a^b\frac{-\cos'(x)}{\cos(x)}dx\\&=-\left.\log(|\cos(x)|)\right|_{x=a}^{x=b} .\end{aligned}\]
For \(a,b\in\mathbb{R}\) holds \[\begin{aligned} \int_a^b\frac{x}{x^2+1}dx=&\frac12\int_a^b\frac{2x}{x^2+1}dx\\=&\frac12\int_a^b\frac{(x^2+1)'}{x^2+1}dx\\=&\frac12\left.\log(|x^2+1|)\right|_{x=a}^{x=b} .\end{aligned}\]
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