5.4 Substitution

5.4.1 Previously Studied Techniques

We begin this section by quickly reviewing the integration techniques that we have used so far.

The three most basic techniques to compute antiderivatives

Thorough knowledge of the derivatives of elementary functions.

Rewrite and simplify, which includes
1. adjust by a constant factor;
2. algebraic manipulations;
3. use of trigonometric identities;
4. use of basic properties of exponential and logarithmic functions.

Identify \(f\), \(g\), and \(g’\) to see the presence of the chain rule, \((f\circ g)'(x) = f'(g(x)) g'(x)\).

Example 1. (Basic knowledge of derivatives of elementary functions. ) Evaluate

(a) \(\displaystyle \int_1^8 \sqrt[3]{x^2}\,dx\); (b) \(\displaystyle \int_{-\pi/6}^{\pi/4}\cos 2x\,dx\); (c) \(\displaystyle \int_{-1}^{\sqrt{3}}\frac{1}{1+t^2}\,dt\).

Solution. (a) Since \(f(x)= \sqrt[3]{x^2}\) is continuous everywhere, and since \(\displaystyle F(x)= \frac{3}{5}x^{5/3}\) is an antiderivative of \(f\) on \(\mathbb{R}\), it follows that we can apply TFTCP2 to obtain
\[\int_1^8 \sqrt[3]{x^2}\,dx = \frac{3}{5}x^{5/3}\Big|_1^8 = \frac{3}{5}\cdot 8^{5/3}-\frac{3}{5}\cdot 1^{5/3}= \frac{3}{5}(2^5-1).\]

(b) Here \(\displaystyle F(x) = \frac{1}{2}\sin 2x\) is an antiderivative of \(f(x) =\cos 2x\). Thus
\[\int_{-\pi/6}^{\pi/4}\cos 2x\,dx = \frac{1}{2}\sin 2x \Big|_{-\pi/6}^{\pi/4} =
\frac{1}{2}\sin \left [2\left (-\frac{\pi}{6}\right )\right ] –
\frac{1}{2}\sin \left [2\left (\frac{\pi}{4}\right )\right ] = \frac{1}{4}\left (2 +\sqrt{3}\right).\]

(c)
\[ \int_{-1}^{\sqrt{3}}\frac{1}{1+t^2}\,dt = \arctan t\Big|_{-1}^{\sqrt{3}} = \frac{\pi}{3}-\left (-\frac{\pi}{4}\right ) = \frac{7\pi}{12}.\]

Example 2. (Rewrite and simplify.) Evaluate the indefinite integral.

\(\displaystyle \int\frac{5+x^2}{1+x^2}\,dx\);
\(\displaystyle \int\tan^2 3x\,dx\);
\(\displaystyle \int\frac{e^{4x}-e^{-4x}}{e^{2x}+e^{-2x}}\,dx\);
\(\displaystyle \int \frac{\sin \pi x}{\cos^2\pi x}\,dx\);
\(\displaystyle \int\frac{\sin 6x}{\sin^3 3x}\,dx\);
\(\displaystyle \int\frac{(x+\sqrt{x})^2}{\sqrt[3]{x^2}}\,dx\).

Solution.
(a) \(\displaystyle \int\frac{5+x^2}{1+x^2}=\int\frac{4+1+x^2}{1+x^2}\,dx
=\int\left(\frac{4}{1+x^2}+ 1\right)\,dx = 4\arctan x+x+C\).

(b) \(\displaystyle \int\tan^2 3x\,dx = \int\left(\sec^2 3x – 1\right)\,dx
= \frac{1}{3}\tan 3x – x + C\).

(c)
\begin{align*}
\int\frac{e^{4x}-e^{-4x}}{e^{2x}+e^{-2x}}\,dx
&=\int\frac{(e^{ 2x})^2-(e^{-2x})^2}{e^{2x}+e^{-2x}}\,dx
=\int\frac{(e^{ 2x}-e^{ 2x})(e^{ 2x}+e^{ -2x})}{e^{2x}+e^{-2x}}\,dx =\\
&=\int (e^{2x} – e^{-2x} )\,dx \underset{?}{=} \frac{1}{2}(e^{2x}+e^{-2x}) + C.
\end{align*}

(d) \(\displaystyle \int \frac{\sin \pi x}{\cos^2\pi x}\,dx
=\int \frac{\sin \pi x}{\cos\pi x}\frac{1}{\cos \pi x}\,dx
=\int \tan \pi x \sec \pi x\,dx = \frac{1}{\pi}\sec \pi x +C\).

(e)

\begin{align*}
\int\frac{\sin 6x}{\sin^3 3x}\,dx
&=\int\frac{\sin (2\cdot 3x)}{\sin^3 4x }\,dx
=\int\frac{2\sin 3x \cos 3x}{\sin^3 3x}\,dx =\int\frac{2\cos 3x }{\sin^2 3x}\,dx = \\
&=\int 2\csc 3x \cot 3x \,dx =- \frac{2}{3}\csc 3x + C.
\end{align*}

(f)
\begin{align*}
\int\frac{\sin 6x}{\sin^3 3x}\,dx
&=\int\frac{\sin (2\cdot 3x)}{\sin^3 4x }\,dx
=\int\frac{2\sin 3x \cos 3x}{\sin^3 3x}\,dx =\int\frac{2\cos 3x }{\sin^2 3x}\,dx = \\
&=\int 2\csc 3x \cot 3x \,dx =- \frac{2}{3}\csc 3x + C.
\end{align*}

Example 3. (Apply the chain rule.) Evaluate

\(\displaystyle \int 3(x^2+2x)^2\,(x+1)\,dx \vphantom{\int_{\sqrt{e}}^{e^2} \frac{1}{x\ln^3 \sqrt[3]{x^2}}\,dx} \);
\(\displaystyle \int \frac{1}{\sqrt{t}} e^{\sqrt{t}} \,dt\);
\(\displaystyle \int_{\sqrt{e}}^{e^2} \frac{1}{x\ln^3 \sqrt[3]{x^2}}\,dx\);
\(\displaystyle \int \frac{ 9\arcsin^3 x}{\sqrt{1-x^2}}\,dx\);
\(\displaystyle \int \frac{3 + 5x}{\sqrt{1-x^2}}\,dx\vphantom{\int_{\sqrt{e}}^{e^2} \frac{1}{x\ln^3 \sqrt[3]{x^2}}\,dx}\);
\(\displaystyle \int 2^{\sin^2x}\sin 2x \,dx \vphantom{ \int \frac{ 9\arcsin^3 x}{\sqrt{1-x^2}}\,dx}\).

Solution. (a) Here, \(g(x)= x^2+2x\), \( g'(x)= 2x+2 = 2(x+1)\), and \( f'(g(x))=3x^2\). Thus \(f(x)=x^3\), and
\begin{align*}
\int 3(x^2+2x)^2\,(x+1)\,dx&= \color{red}{\pmb{\frac{1}{2}}}\int 3(x^2+2x)^2\;\color{red}{\pmb{2}}\;(x+1)\,dx
=\frac{1}{2}(x^2+2x)^3 +C.
\end{align*}

(b) Let \(\displaystyle g(t)= \sqrt{t} \), \(\displaystyle g'(t)= \frac{1}{2}\frac{1}{\sqrt{t}} \), and \(f'(t)=e^{t}\). Then \(f(t)=e^t+C\), and we have
\begin{align*}
\int \frac{1}{\sqrt{t}} e^{\sqrt{t}} \,dt
&= \color{red}{\pmb{2}}\int e^{\sqrt{t}}\,\cdot\color{red}{\pmb{\frac{1}{2}}}\;\frac{1}{\sqrt{t}} \,dt
= \color{red}{\pmb{2}} e^{\sqrt{t}} +C.
\end{align*}

(c) First, we rewrite the expression involving the natural logarithm:
\[\ln^3 \sqrt[3]{x^2} = \left (\ln x^{2/3}\right )^3 = \left (\frac{2}{3}\ln \,| x\,|\right )^3 = \frac{8}{27}\ln^3\,| x\,|.\]
Thus
\[\int \frac{1}{x\ln^3 \sqrt[3]{x^2}}\,dx
= \int \frac{1}{x\left (\frac{2}{3}\ln x\right )^3}\,dx
=\frac{27}{8} \int \left (\ln \,| x\,|\right )^{-3}\frac{1}{x}\,dx.\]
Thus
\[g(x) = \ln \,| x\,|, \Longrightarrow g'(x) = \frac{1}{x}; \;\;\; f'(x) = x^{-3}, \Longrightarrow f(x) = -\frac{1}{2}x^{-2}+C.\]
Then
\[\int \frac{1}{x\ln^3 \sqrt[3]{x^2}}\,dx
=\frac{27}{8} \int \left (\ln \,| x\,|\right )^{-3}\frac{1}{x}\,dx
=-\frac{27}{16}\ln^{-2}\,| x\,|+C.\]
And
\begin{align*}
\int_{\sqrt{e}}^{e^2} \frac{1}{x\ln^3 \sqrt[3]{x^2}}\,dx
&= -\frac{27}{16}\ln^{-2}x\, \Big|_{\sqrt{e}}^{e^2}
=-\frac{27}{16}\left (\ln^{-2}e^2 – \ln^{-2}e^{1/2}\right )
= -\frac{27}{16}\left (2^{-2} -\left (\frac{1}{2}\right )^{-2}\right ) = \\
&=\frac{27}{16}\cdot\frac{15}{4}.
\end{align*}

(d) \(\displaystyle g(x) = \arcsin x, \Longrightarrow g'(x) =\frac{1}{\sqrt{1-x^2}}; \;\;\; f'(x) = x^3, \Longrightarrow f(x) =\frac{1}{4} x^{4} + C\). Hence
\[\int \frac{ 9\arcsin^3 x}{\sqrt{1-x^2}}\,dx = \frac{9}{4}\arcsin^4 x+ C.\]

(e) Since \(\displaystyle \int \frac{3 + 5x}{\sqrt{1-x^2}}\,dx =
3\int \frac{1 }{\sqrt{1-x^2}}\,dx + 5 \int \frac{ x }{\sqrt{1-x^2}}\,dx\), we evaluate each integral separately.
\[ 3\int \frac{1}{\sqrt{1-x^2}}\,dx = 3\arcsin x +C_0;\]

\[5 \int \frac{x}{\sqrt{1-x^2}}\,dx = 5 \int x (1-x^2)^{-1/2}\,dx\]
\[ g(x) = 1-x^2, \Longrightarrow g'(x) = -2x; \;\;\; f'(x) = x^{-1/2}, \Longrightarrow f(x) =\frac{2}{3} x^{3/2} + C_1. \Longrightarrow\]
\[\Longrightarrow \int \frac{ x }{\sqrt{1-x^2}}\,dx=
5\cdot \, \pmb{\left (\!-\frac{1}{2}\!\right )} \int \frac{ \pmb{-2} \, x }{\sqrt{1-x^2}}\,dx = -\frac{5}{2}\cdot\frac{2}{3}(1-x^2)^{3/2}+C_1 =
-\frac{5}{3}(1-x^2)\sqrt{1-x^2}+C_1. \]
Thus
\[\int \frac{3 + 5x}{\sqrt{1-x^2}}\,dx = 3\arcsin x -\frac{5}{3}(1-x^2)\sqrt{1-x^2}+C.\]

(f) Since \(\displaystyle \int 2^{\sin^2x}\sin 2x \,dx = \int 2^{\sin^2x}2\sin x\cos x \,dx\),
\[g(x) = \sin^2x, \Longrightarrow g'(x) = 2\sin x \cos x; \;\;\; f'(x) = 2^x = e^{\ln 2^x} = e^{x\ln 2}, \Longrightarrow f(x) = \frac{1}{\ln 2}e^{x\ln 2}+C.\]
Then
\[\int 2^{\sin^2x}\sin 2x \,dx = \int 2^{\sin^2x}2\sin x\cos x \,dx = \frac{1}{\ln 2}e^{ \sin^2x\cdot\ln 2}+C= \frac{1}{\ln 2}2^{\sin^2x}+C.\]

5.4.2 Exercises

In problems 1—12, evaluate the integral.

\(\displaystyle \int x^3 (2 + x^4)^2\,dx \vphantom{ \int \frac{\cos^2\pi x}{\sin^4\pi x}\,dx}\).
\(\displaystyle \int t^2\sec^2 (t^3)\,dt\).
\(\displaystyle \int_{1/2}^{2/3} \cos \pi x \sin^3\pi x \,dx \vphantom{\int \left (\frac{2^{\sqrt{t}}}{\sqrt{t}}+ e^{e^t+t} \right )\,dt} \vphantom{\int_{-1}^{1/\sqrt{3}} \frac{\arctan^2 x}{1+x^2}\,dx}\).
\(\displaystyle \int e^{2w} \cos e^{2w} \,dw \vphantom{\int \frac{2+ 3x + 4x^2}{1+x^2}\,dt}\).
\(\displaystyle \int_{\sqrt{e}}^{e^2} \frac{1}{x\ln^3 x}\,dx \vphantom{ \int \frac{\cos^2\pi x}{\sin^4\pi x}\,dx}\).
\(\displaystyle \int_{-\sqrt{3}/2}^{1/2} \frac{\arcsin x}{\sqrt{1-x^2}} \,dx \vphantom{\int \left (\frac{2^{\sqrt{t}}}{\sqrt{t}}+ e^{e^t+t} \right )\,dt}\).
\(\displaystyle \int_{-1}^{1/\sqrt{3}} \frac{\arctan^2 x}{1+x^2}\,dx\);
\(\displaystyle \int e^{\cos^2 \pi x}\sin 2\pi x \,dx \vphantom{\int \frac{2+ 3x + 4x^2}{1+x^2}\,dt}\).
\(\displaystyle \int \frac{\cos^2\pi x}{\sin^4\pi x}\,dx\).
\(\displaystyle \int \left (\frac{2^{\sqrt{t}}}{\sqrt{t}}+ e^{e^t+t} \right )\,dt\).
\(\displaystyle \int \frac{\cos 2\pi x}{\cos \pi x – \sin \pi x} \, dx \vphantom{\int_{-1}^{1/\sqrt{3}} \frac{\arctan^2 x}{1+x^2}\,dx}\).
\(\displaystyle \int \frac{2+ 3x + 4x^2}{1+x^2}\,dt\).

5.4.3 Substitution

Theorem 1. If \(u=g(x)\) is a differentiable function whose range is contained on the open interval \(I\), and \(f\) is continuous on \(I\), then
\begin{equation}\tag{5.6}
\int f(g(x))g'(x)\,dx = \int f(u)\,du.
\end{equation}

Proof.
Let \(F\) be an antiderivative of \(f\) on \(I\), then, by the chain rule,
\[\int f(g(x))g'(x)\,dx = F(g(x)) + C\;\;\; \text{and}\;\;\;\int f(u)\,du. = F(u) + C,\]
and the result follows.

Key Steps When Applying the Substitution Rule

Given an indefinite integral, \(\displaystyle \int f(\color{red}{\pmb{g(x)}})\,\color{red}{\pmb{g'(x)}}\,dx\),

identify \(u=g(x)\), and \(du = g'(x)\,dx\),

it may be necessary to rewrite the integrand to identify these two terms;

note that \(g'(x)\,dx\) is in the numerator of the integrand.
make the substitution into the integral and obtain the simpler integral \(\displaystyle \int f(u)\,du\),
evaluate the indefinite integral \(\displaystyle F(u)=\int f(u)\,du\),
then, \(\displaystyle \int f(g(x))g'(x)\,dx = F(g(x))+C\).

Example 1. Evaluate \(\displaystyle \int \frac{ \cos \pi x}{\sqrt{1-\sin^2\pi x}}\,dx\).

Solution.

identify \(u=g(x)=\sin \pi x\), and \(du = g'(x)\,dx = \pi \cos \pi x \,dx\),
Note that it is necessary to adjust the integral by the constant factor \(\displaystyle \frac{1}{\pi} \):
\[\int \frac{ \cos \pi x}{\sqrt{1-\sin^2\pi x}}\,dx = \frac{1}{\pi}\int \frac{\pi \cos \pi x}{\sqrt{1-\sin^2\pi x}}\,dx.\]
make the substitution into the integral: \(\displaystyle \frac{1}{\pi} \int \frac{1}{\sqrt{1-u^2}}\,du\),
evaluate the indefinite integral: \(\displaystyle F(u)=\frac{1}{\pi}\int \frac{1}{\sqrt{1-u^2}}\,du =\frac{1}{\pi}\arcsin u+C\),
then, replacing \(u = g(x)\), we have
\[ \int \frac{ \cos \pi x}{\sqrt{1-\sin^2\pi x}}\,dx = F(g(x))+C
= \frac{1}{\pi}\arcsin (\sin \pi x) + C.\]

Note. For \(\displaystyle -\frac{1}{2}<x<\frac{1}{2}\), the answer simplifies to \(\pi x +C\) (explain why).

Example 2. Evaluate \(\displaystyle \int \frac{1}{x \sqrt[4]{\ln^3x^5}}\,dx\).

Solution.

identify \(u=g(x)\) and \(du = g'(x)\,dx\).

First, rewrite the integrand,
\[\sqrt[4]{\ln^3x^5} =(5\ln x)^{3/4} =5^{3/4}\left (\ln x \right )^{3/4}.\]
Thus
\[u = \ln x\;\;\; \text{and} \;\;\; du = g'(x)\,dx = \frac{1}{x}\,dx.\]

make the substitution into the integral
\[ \int \frac{1}{x \sqrt[4]{\ln^3x^5}}\,dx = \int \frac{1}{x\cdot 5^{3/4}\left (\ln x \right )^{3/4}}\,dx = \frac{1}{5^{3/4}}\int \frac{1}{ u^{3/4}}\,du = \frac{1}{5^{3/4}}\int u^{-3/4}\,du.\]
evaluate the indefinite integral
\[\frac{1}{5^{3/4}}\int u^{-3/4}\,du = \frac{1}{5^{3/4}} \cdot 4 u^{1/4}+C = \frac{4}{5^{3/4}} \sqrt[4]{u}+C.\]
replace \(u = g(x)\)
\[\int \frac{1}{x \sqrt[4]{\ln^3x^5}}\,dx = \frac{4}{5^{3/4}}\, \sqrt[4]{\ln x}+C.\]

5.4.4 Exercises

Evaluate the following indefinite integrals by means of the indicated substitution:

\(\displaystyle \int x \sqrt[3]{x^2+1}\, dx\), \( u= x^2+1\).
\(\displaystyle \int \frac{x}{1+x^4}\,dx\), \(u = x^2\).
\(\displaystyle \int e^{3x} \tan e^{3x} \,dx\), \(u=\cos e^{3x}\).
\(\displaystyle \int \frac{1}{\sqrt{t}}\tan \sqrt{t}\sec \sqrt{t}\,dt\), \(u = \sqrt{t}\).
\(\displaystyle \int \frac{\arctan 2t}{1+4t^2} \,dt\), \(u = \arctan 2t\).
\(\displaystyle \int \frac{ e^{\cos \pi x}\sin \pi x}{\sqrt{1-e^{2 \cos \pi x}}}\, dx\), \(u = e^{\cos \pi x}\).
\(\displaystyle \int \frac{\sin 2\pi x}{1+ \cos^4 \pi x} \, dx\), \(u = \cos^2 \pi x\).
\(\displaystyle \int \frac{1}{x\, \sqrt{1-\ln^2x^3}} \, dx\), \(u = \ln x^3\).

In problems 9—14, evaluate the indefinite integral via an appropriate substitution.

\(\displaystyle \int \frac{x^2}{\sqrt[3]{x^3+1}} \,dx\).
\(\displaystyle \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \,dx\).
\(\displaystyle \int e^{3x} \cot e^{3x} \,dx\).
\(\displaystyle \int \frac{\cot \sqrt[3]{x}\csc^2\sqrt[3]{x} }{\sqrt[3]{x^2} }\,dx\).
\(\displaystyle \int \frac{1}{x \,\sqrt{1 + \ln^2 x^3}} \,dx\).
\(\displaystyle \int \frac{\sin 4\pi x}{\sqrt{1- \sin^4 2\pi x}} \, dx\).

In problems 15—20, evaluate the definite integral.

\(\displaystyle \int_{1\left/\sqrt{e^3}\right.}^{e^2} \frac{(\ln x)^3}{x} \, dx\)
\(\displaystyle \int_{1\left/\sqrt{3}\right.}^{\sqrt{3}} \frac{1+\arctan x}{1+x^2} \, dx\)
\(\displaystyle \int_{1/4}^{1/3} \frac{\sin \pi x}{\sqrt{1-\cos^2 \pi x}} \, dx\)
\(\displaystyle \int_{-\sqrt{3}/2}^{1/2} \frac{\arcsin x}{\sqrt{1-x^2}} \, dx\)
\(\displaystyle \int _{\sqrt{\pi} /2}^{\sqrt{\pi /3}} x \tan^2 x^2 \sec^2 x^2 \,dx\)
\(\displaystyle \int _{ \pi /6}^{ \pi /3}e^{\sin^2 x}\sin 2 x\,dx\)

Previous Section

Next Section