\(u\)-Substitution

Along with integration by parts, the \(u\)-substitution is an integration technique that is frequently used for integrals that cannot be directly solved. The procedure is as follows:

• (i) Find the term to be substituted for, and let that be \(u.\)

• (ii) Find \(du\) \((\)in terms of \(dx).\)

• (iii) Substitute \(u\) and \(du\) into the expression.

• (iv) Integrate with respect to \(u,\) and then substitute \(x\) back into the result.

For instance, say we have an integral of the form \[\int f(x)f'(x)\, dx.\] The process of solving this using \(u\)-substitution is as follows:

• (i) First we find the part to replace with \(u.\) In this case letting \(u=f(x)\) would be appropriate.

• (ii) Then we find \(du\) by differentiating both sides of \(u=f(x):\) \[du=f'(x)\, dx.\]

• (iii) Now, replacing \(x\) with \(u\) and \(dx\) with \(du\) gives \[\int f(x)f'(x)\, dx=\int u\, du.\]

• (iv) Finally, we integrate with respect to \(u,\) and substitute \(x\) back into the result to obtain \[\int u\, du=\frac{1}{2}u^2+C=\frac{1}{2}\big(f(x)\big)^2+C,\] where \(C\) is the constant of integration.

Observe that differentiating the result using differentiation of composite functions gives \(f(x)f'(x).\) As you get used to \(u\)-substitution, you will find out that integration via \(u\)-substitution is the exact opposite of differentiation of composite functions, but let's just stick to the basics for now.

For starters, it might not be so easy to determine for which part of the expression should be substituted. So, for our first step, we will start with problems where the substitution is given. As you walk through these examples, you will be gaining some sense in deciding which part of the integral to substitute for.

Using the substitution \(u=x^3+1,\) find \(\displaystyle{\int x^2\sqrt{x^3+1}~dx.}\)

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From \(u=x^3+1,\) we have \(du=3x^2dx.\) Hence the given expression is equivalent to \[\begin{align} \int x^2\sqrt{x^3+1}\,dx&=\int\frac{1}{3}\sqrt{u}\,du\\ &=\frac{2}{9}u^{\frac{3}{2}}+C\\ &=\frac{2}{9}\big(x^3+1\big)\sqrt{x^3+1}+C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

Using the substitution \(u=x^3,\) find \(\displaystyle{\int x^2e^{x^3}dx.}\)

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From \(u=x^3,\) we have \(du=3x^2dx.\) Hence the given expression is equivalent to \[\int x^2e^{x^3}dx=\int\frac{1}{3}e^udu=\frac{1}{3}e^{x^3}+C,\] where \(C\) is the constant of integration. \(_\square\)

Using the substitution \(u=x+3,\) find \(\displaystyle{\int x(x+3)^9dx.}\)

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From \(u=x+3,\) we have \(du=dx.\) Hence the given expression is equivalent to \[\begin{align} \int x(x+3)^9dx&=\int(u-3)u^9du\\ &=\int \big(u^{10}-3u^9\big)\,du\\ &=\frac{1}{11}u^{11}-\frac{3}{10}u^{10}+C\\ &=\frac{1}{11}(x+3)^{11}-\frac{3}{10}(x+3)^{10}+C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

Using the substitution \(u=\sin x,\) find \(\displaystyle{\int \sin^5 x\cos x\, dx.}\)

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From \(u=\sin x,\) we have \(du=\cos xdx.\) Hence the given expression is equivalent to \[\begin{align} \int \sin^5 x\cos x\, dx&=\int u^5du\\ &=\frac{1}{6}u^6+C\\ &=\frac{1}{6}\sin^6 x+C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

When it comes to integration using \(u\)-substitution of definite integrals, we need to keep one more thing in mind: the integration interval must be changed to the interval of \(u\) that corresponds to the given interval of \(x.\) Let's try the definite integrals of the functions we have integrated above.

Using the substitution \(u=x^3+1,\) find \(\displaystyle{\int_0^2 x^2\sqrt{x^3+1}~dx.}\)

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From \(u=x^3+1,\) we have \[\begin{align} du&=3x^2dx\\ x&=0\rightarrow u=1\\ x&=2\rightarrow u=9. \end{align}\] Hence the given expression is equivalent to \[\begin{align} \int_0^2 x^2\sqrt{x^3+1}\,dx&=\int_1^9\frac{1}{3}\sqrt{u}\,du\\ &=\left.\frac{2}{9}u^{\frac{3}{2}}\right|_1^9\\ &=\frac{52}{9}.\ _\square \end{align}\]

Using the substitution \(u=x^3,\) find \(\displaystyle{\int_0^2 x^2e^{x^3}dx.}\)

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From \(u=x^3,\) we have \[\begin{align} du&=3x^2dx\\ x&=0\rightarrow u=0\\ x&=2\rightarrow u=8. \end{align}\] Hence the given expression is equivalent to \[\int_0^2 x^2e^{x^3}dx=\int_0^8\frac{1}{3}e^udu=\left.\frac{1}{3}e^u\right|_0^8=\frac{1}{3}\big(e^8-1\big).\ _\square\]

Using the substitution \(u=x+3,\) find \(\displaystyle{\int_{-3}^0 x(x+3)^9dx.}\)

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From \(u=x+3,\) we have \[\begin{align} du&=dx\\ x=-3&\rightarrow u=0\\ x=0&\rightarrow u=3. \end{align}\] Hence the given expression is equivalent to \[\begin{align} \int_{-3}^0 x(x+3)^9dx&=\int_0^3(u-3)u^9du\\ &=\int_0^3\big(u^{10}-3u^9\big)du\\ &=\left[\frac{1}{11}u^{11}-\frac{3}{10}u^{10}\right]_0^3\\ &=-\frac{3^{11}}{110}.\ _\square \end{align}\]

Using the substitution \(u=\sin x,\) find \(\displaystyle{\int_0^{\frac{\pi}{2}} \sin^5 x\cos x\,dx.}\)

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From \(u=\sin x,\) we have \[\begin{align} du&=\cos x\,dx\\ x&=0\rightarrow u=0\\ x&=\frac{\pi}{2}\rightarrow u=1. \end{align}\] Hence the given expression is equivalent to \[\int_0^{\frac{\pi}{2}} \sin^5 x\cos x\,dx=\int_0^1 u^5du=\left.\frac{1}{6}u^6\right|_0^1=\frac{1}{6}.\ _\square\]

\(u\)-substitution is a great way to simplify integrals. It is a technique used in many other forms of integration such as integration by parts and the infamous trig sub.

\(u\)-substitutions take two general forms, where \(f(x)=u\) or \(f(u)=x\). Note that the chain rule for differentiation is basically equivalent to \(u\)-sub for integrals.

Evaluate \( \displaystyle \int x\ln x\, dx\).

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Let \(x=e^u\) and \(dx=e^u\,du\). Then think of the \(dx\) as a "variable" and substitute it for \(dx=e^u\, du:\) \[ \int e^u \ln e^u (e^u\,du)= \int e^{2u} u\,du.\] Doing a \(w\)-sub (the same thing as a \(u\)-sub but with a different variable), we have \[\begin{align} w&=2u\\ dw&=2\, du\\ du&=\frac{dw}{2}\\ \Rightarrow \int e^{2u} u\,du &=\int e^{w}\left( \frac{w}{2}\right) \left(\frac{dw}{2}\right)\\ &=\frac{1}{4} \int we^w\,dw. \qquad (1) \end{align}\] Now, integrating by parts, \(\displaystyle \int a\,db=ab-\displaystyle \int b\,da.\) Note that \(db\) and \(da\) don't mean differentiating with respect to \(a\) and \(b\), but \(da\) is the derivative of \(a\) and \(db\) is the derivative of \(b\). So, here we let \(a=w\) and \(db=e^w\), then \(da=1\) and \(b=e^w\) \((\)both the integral and derivative of \(e^w\) are \(e^w).\)

Hence, \((1)\) can be rewritten as \[\begin{align} \frac{1}{4} \int we^w dw &=\frac{1}{4}we^w-\frac{1}{4}\displaystyle \int e^w dw\\ &=\frac{1}{4}(we^w-e^w). \qquad (2) \end{align}\] For the last steps of our \(w\)- and \(u\)-substitutions, we must re-substitute for \(w\) and \(u:\) \(w=2u\) and \(u=\ln(x)\). Then we can rewrite \((2)\) as \[\begin{align} \frac{1}{4}\left(we^w-e^w\right) &=\frac{1}{4}\left(2ue^{2u}-e^{2u}\right)\\ &=\frac{1}{4}\big(2(\ln x)e^{2\ln x}-e^{2\ln x}\big)\\ &=\frac{1}{4}\left(2x^2\ln x-x^2\right)\\ &=\frac{1}{4}x^2(2\ln x-1). \ _\square \end{align}\]

Now we will discuss a particular but frequently used form of \(u\)-substitution, which is the \(\ln\big\lvert f(x)\big\rvert\) form. Consider the integral of the form \[\int \dfrac{f'(x)}{f(x)}dx. \qquad (1)\] We substitute \(u=f(x)\) so that \(du=f'(x)\,dx\). Then \((1)\) is equivalent to \[\int\dfrac{1}{u}\,du=\ln \lvert u\rvert+C=\ln \big\lvert f(x)\big\rvert+C,\] where \(C\) is the constant of integration. Hence we can easily compute the integrals of form \((1)\), or we can manipulate expressions to turn them into that form.

Evaluate \(\displaystyle{\int\frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx.}\)

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Let \(u=e^x+e^{-x}.\) Then we have \(du=(e^x-e^{-x})\,dx,\) and it follows that \[\int\frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx=\int\frac{du}{u}=\ln\lvert u\rvert+C=\ln\big\lvert e^x+e^{-x}\big\rvert+C,\] where \(C\) is the constant of integration. \(_\square\)

Evaluate \(\displaystyle{\int\frac{2x+3}{x^2+3x-4}\,dx.}\)

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Let \(u=x^2+3x-4.\) Then we have \(du=(2x+3)\,dx,\) and it follows that \[\begin{align} \int\frac{2x+3}{x^2+3x-4}\,dx&=\int\frac{du}{u}\\ &=\ln\lvert u\rvert+C\\ &=\ln\big\lvert x^2+3x-4\big\rvert+C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

Evaluate \(\displaystyle \int \cot x\,dx\).

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We have \[\int \cot x\,dx=\int \dfrac{\cos x}{\sin x}\,dx=\int\dfrac{(\sin x)'}{\sin x}dx=\ln\lvert\sin x\rvert+C,\] where \(C\) is the constant of integration. \(_\square\)

Evaluate \(\displaystyle \int \dfrac{1}{\ln x^x}dx\).

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We have \[\int \dfrac{1}{\ln x^x}dx=\int \dfrac{1}{x\ln x}dx=\int \dfrac{\frac{1}{x}}{\ln x}dx=\int \dfrac{(\ln x)'}{\ln x}dx=\ln\lvert\ln x\rvert+C,\] where \(C\) is the constant of integration. \(_\square\)