6.7 integration by substitution homework answer key

Section 6.7. Integration by substitution

Calculus Tutorials

Computing integrals by substitution.

Many integrals are most easily computed by means of a change of variables, commonly called a $u$-substitution .

Let’s compute $\displaystyle\int\! 2x(x^2-1)^4\, dx$ by making the substitution \begin{eqnarray*} u&=&x^2-1\\ du&=&2x\, dx. \end{eqnarray*} Then \[ \int 2x(x^2-1)^4\, dx=\int (x^2-1)^4(2x\, dx)=\int u^4\, du=\frac{u^5}{5}+C=\frac{(x^2-1)^5}{5}+C.\] We may check this result by differentiating using the Chain Rule: \[\frac{d}{dx}\left(\frac{(x^2-1)^5}{5}+C\right)=\frac{5(x^2-1)^4}{5}(2x) =2x(x^2-1)^4.\qquad\qquad \surd\]

The substitution method amounts to applying the Chain Rule in reverse:

To compute $\displaystyle\int\! f(g(x))g'(x)\, dx$, we let \begin{eqnarray*} u&=&g(x)\\ du&=&g'(x)\, dx. \end{eqnarray*} Then \[\int f(g(x))g'(x)\, dx=\int f(u)\, du=F(u)=F(g(x))\] where $F$ is an antiderivative of $f$.

To compute $\displaystyle\int\! \sin (2x)\cos (2x)\, dx$, let \begin{eqnarray*} u&=&\sin (2x)\\ du&=&2\cos (2x)\, dx. \end{eqnarray*} Then \[\int \sin (2x)\cos (2x)\, dx=\int\frac{1}{2}\sin (2x)[2\cos (2x)\, dx]=\int \frac{1}{2}u\, du=\frac{1}{4}u^2+C=\frac{1}{4}\sin^2 (2x)+C.\]

With practice, you will often be able to write down the result immediately.

We can evaluate $\displaystyle\int\! \frac{dx}{(4x-3)^2}$ by letting \begin{eqnarray*} u&=&4x-3\\ du&=&4\, dx\quad\longrightarrow\quad dx=\frac{1}{4}\, du. \end{eqnarray*} Then \[\int \frac{dx}{(4x-3)^2}=\int \frac{\frac{1}{4}\, du}{u^2}=-\frac{1}{4u}+C=\frac{-1}{4(4x-3)}+C.\]

It is not always apparent until you try it whether or not a substitution will work.

To compute $\displaystyle\int\! x\sqrt{x-3}\, dx$, we will try \begin{eqnarray*} u&=&x-3\quad\longrightarrow\quad x=u+3\\ du&=&dx. \end{eqnarray*} So \begin{eqnarray*} \int x\sqrt{x-3}\, dx&=&\int (u+3)\sqrt{u}\, du=\int \left(u^{3/2}+3u^{1/2}\right)\, du\\ &=&\frac{2}{5}u^{5/2}+2u^{3/2}+C=\frac{2}{5}(x-3)^{5/2}+2(x-3)^{3/2}+C. \end{eqnarray*}

We can also compute a definite integral using a substitution.

Let’s evaluate $\displaystyle\int^2_0\! xe^{x^2}\, dx$. Let \begin{eqnarray*} u&=&x^2\\ du&=&2x\, dx. \end{eqnarray*} First, we will compute the indefinite integral: \[\int xe^{x^2}\, dx=\int \left(\frac{1}{2}e^{x^2}\right)(2x\, dx)=\int\frac{1}{2}e^u\, du=\frac{1}{2}e^u+C=\frac{1}{2}e^{x^2}+C.\] Now we have two approaches for the definite integral:

Thus, we find that \[\int^2_0 xe^{x^2}\, dx=\frac{1}{2}(e^4-1).\]

Approach 2 works provided certain conditions on $f$ and $g$ are met: \[\int^b_a f(g(x))\, dx=\int^{g(b)}_{g(a)} f(u)\, du\] if

• $g’$ is continuous on $[a,b]$.
• $f$ is continuous on the set of values taken by $g$ on $[a,b]$.

Substitutions are useful or necessary for a huge range of integrals. You will find yourself either implicitly or explicitly using a substitution in virtually every integral you compute!

Key Concepts

To compute $\displaystyle\int\! f(g(x))g'(x)\, dx$, we let $u = g(x)$ $du = g'(x) dx$.

Then $\displaystyle{\int f(g(x))g'(x)\, dx = \int f(u)\, du = F(u) = F(g(x))}$ where $F$ is an antiderivative of $f$.

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Answer Key 6.7

• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{20x^4}{4x^3}&+&\dfrac{x^3}{4x^3}&+&\dfrac{2x^2}{4x^3} \\ \\ 5x&+&\dfrac{1}{4}&+&\dfrac{1}{2x} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{5x^4}{9x}&+&\dfrac{45x^3}{9x}&+&\dfrac{4x^2}{9x} \\ \\ \dfrac{5}{9}x^3&+&5x^2&+&\dfrac{4}{9}x \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{20n^4}{10n}&+&\dfrac{n^3}{10n}&+&\dfrac{40n^2}{10n} \\ \\ 2n^3&+&\dfrac{n^2}{10}&+&4n \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{3k^3}{8k}&+&\dfrac{4k^2}{8k}&+&\dfrac{2k}{8k} \\ \\ \dfrac{3}{8}k^2&+&\dfrac{k}{2}&+&\dfrac{1}{4} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{12x^4}{6x}&+&\dfrac{24x^3}{6x}&+&\dfrac{3x^2}{6x} \\ \\ 2x^3&+&4x^2&+&\dfrac{x}{2} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{5p^4}{4p}&+&\dfrac{16p^3}{4p}&+&\dfrac{16p^2}{4p} \\ \\ \dfrac{5}{4}p^3&+&4p^2&+&4p \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{10n^4}{10n^2}&+&\dfrac{50n^3}{10n^2}&+&\dfrac{2n^2}{10n^2} \\ \\ n^2&+&5n&+&\dfrac{1}{5} \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{ccccc} \dfrac{3m^4}{9m^2}&+&\dfrac{18m^3}{9m^2}&+&\dfrac{27m^2}{9m^2} \\ \\ \dfrac{m^2}{3}&+&2m&+&3 \end{array}[/latex]

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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6.8E: Exercises for Section 6.7

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In exercises 1 - 3, find the derivative \(\dfrac{dy}{dx}\).

1) \(y=\ln(2x)\)

2) \(y=\ln(2x+1)\)

3) \(y=\dfrac{1}{\ln x}\)

In exercises 4 - 5, find the indefinite integral.

4) \(\displaystyle ∫\frac{dt}{3t}\)

5) \(\displaystyle ∫\frac{dx}{1+x}\)

In exercises 6 - 15, find the derivative \(\dfrac{dy}{dx}.\) (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

6) [T] \(y=\dfrac{\ln x}{x}\)

7) [T] \(y=x\ln x\)

8) [T] \(y=\log_{10}x\)

9) [T] \(y=\ln(\sin x)\)

10) [T] \(y=\ln(\ln x)\)

11) [T] \(y=7\ln(4x)\)

12) [T] \(y=\ln\big((4x)^7\big)\)

13) [T] \(y=\ln(\tan x)\)

14) [T] \(y=\ln(\tan 3x)\)

15) [T] \(y=\ln(\cos^2x)\)

In exercises 16 - 25, find the definite or indefinite integral.

16) \(\displaystyle ∫^1_0\frac{dx}{3+x}\)

17) \(\displaystyle ∫^1_0\frac{dt}{3+2t}\)

18) \(\displaystyle ∫^2_0\frac{x}{x^2+1}\, dx\)

19) \(\displaystyle ∫^2_0\frac{x^3}{x^2+1}\,dx\)

20) \(\displaystyle ∫^e_2\frac{dx}{x\ln x}\)

21) \(\displaystyle ∫^e_2\frac{dx}{(x\ln x)^2}\)

22) \(\displaystyle ∫\frac{\cos x}{\sin x}\, dx\)

23) \(\displaystyle ∫^{π/4}_0\tan x\,dx\)

24) \(\displaystyle ∫\cot(3x)\,dx\)

25) \(\displaystyle ∫\frac{(\ln x)^2}{x}\, dx\)

In exercises 26 - 35, compute \(\dfrac{dy}{dx}\) by differentiating \(\ln y\).

26) \(y=\sqrt{x^2+1}\)

27) \(y=\sqrt{x^2+1}\sqrt{x^2−1}\)

28) \(y=e^{\sin x}\)

29) \(y=x^{−1/x}\)

30) \(y=e^{ex}\)

31) \(y=x^e\)

32) \(y=x^{(ex)}\)

33) \(y=\sqrt{x}\sqrt[3]{x}\sqrt[6]{x}\)

34) \(y=x^{−1/\ln x}\)

35) \(y=e^{−\ln x}\)

In exercises 36 - 40, evaluate by any method.

36) \(\displaystyle ∫^{10}_5\dfrac{dt}{t}−∫^{10x}_{5x}\dfrac{dt}{t}\)

37) \(\displaystyle ∫^{e^π}_1\dfrac{dx}{x}+∫^{−1}_{−2}\dfrac{dx}{x}\)

38) \(\dfrac{d}{dx}\left[\displaystyle ∫^1_x\dfrac{dt}{t}\right]\)

39) \(\dfrac{d}{dx}\left[\displaystyle ∫^{x^2}_x\dfrac{dt}{t}\right]\)

40) \(\dfrac{d}{dx}\Big[\ln(\sec x+\tan x)\Big]\)

In exercises 41 - 44, use the function \(\ln x\). If you are unable to find intersection points analytically, use a calculator.

41) Find the area of the region enclosed by \(x=1\) and \(y=5\) above \(y=\ln x\).

42) [T] Find the arc length of \(\ln x\) from \(x=1\) to \(x=2\).

43) Find the area between \(\ln x\) and the \(x\)-axis from \(x=1\) to \(x=2\).

44) Find the volume of the shape created when rotating this curve from \(x=1\) to \(x=2\) around the \(x\)-axis, as pictured here.

45) [T] Find the surface area of the shape created when rotating the curve in the previous exercise from \(x=1\) to \(x=2\) around the \(x\)-axis.

If you are unable to find intersection points analytically in the following exercises, use a calculator.

46) Find the area of the hyperbolic quarter-circle enclosed by \(x=2\) and \(y=2\) above \(y=1/x.\)

47) [T] Find the arc length of \(y=1/x\) from \(x=1\) to \(x=4\).

48) Find the area under \(y=1/x\) and above the \(x\)-axis from \(x=1\) to \(x=4\).

In exercises 49 - 53, verify the derivatives and antiderivatives.

49) \(\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2+1})\Big]=\dfrac{1}{\sqrt{1+x^2}}\)

50) \(\dfrac{d}{dx}\Big[\ln\left(\frac{x−a}{x+a}\right)\Big]=\dfrac{2a}{(x^2−a^2)}\)

51) \(\dfrac{d}{dx}\Big[\ln\left(\frac{1+\sqrt{1−x^2}}{x}\right)\Big]=−\dfrac{1}{x\sqrt{1−x^2}}\)

52) \(\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2−a^2})\Big]=\dfrac{1}{\sqrt{x^2−a^2}}\)

53) \(\displaystyle ∫\frac{dx}{x\ln(x)\ln(\ln x)}=\ln|\ln(\ln x)|+C\)

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .