1.4E: Exercises

Reading Questions

1. What is the main goal of the integration by substitution method?
2. If we choose \(u=g(x)\) for a substitution, what is \(du\) equal to?
3. When is integration by substitution particularly useful? What form of integrand are we often looking for?
4. After performing a \( u \)-substitution for an indefinite integral and finding the antiderivative in terms of \(u\), what is the final step?
5. When performing a \( u \)-substitution for a definite integral \(\displaystyle \int_a^b f(g(x))g^{\prime}(x) \, dx\), what must be done to the limits of integration \(a\) and \(b\)?
6. In Example 2, where \(u=z^2-5\) and \(du=2z \, dz\), but the integral had \(z \, dz\), what "alteration" was made?
7. Is it permissible to have both \(x\) and \(u\) (or \(dx\) and \(du\)) in the integrand after making a substitution?
8. If, after choosing \(u=g(x)\), there are still terms involving \(x\) in the integrand that are not part of \(du\), what algebraic step might be necessary?
9. What is meant by "substitution with resubstitution" as illustrated in Example 5?
10. Why is it critical to change the limits of integration immediately when performing a substitution on a definite integral?

Homework

In exercises 1 - 10, find the antiderivatives for the given functions.

1) \(\cosh(2x+1)\)

Answer

\(\frac{1}{2}\sinh(2x+1)+C\)

2) \(\tanh(3x+2)\)

3) \(x\cosh(x^2)\)

Answer

\(\frac{1}{2}\sinh^2(x^2)+C\)

4) \(3x^3\tanh(x^4)\)

5) \(\cosh^2(x)\sinh(x)\)

Answer

\(\frac{1}{3}\cosh^3(x)+C\)

6) \(\tanh^2(x)\text{sech}^2(x)\)

7) \(\dfrac{\sinh(x)}{1+\cosh(x)}\)

Answer

\(\ln(1+\cosh(x))+C\)

8) \(\coth(x)\)

9) \(\cosh(x)+\sinh(x)\)

Answer

\(\cosh(x)+\sinh(x)+C\)

10) \((\cosh(x)+\sinh(x))^n\)

In exercises 11 - 17, find the antiderivatives for the functions.

11) \(\displaystyle \int\frac{dx}{4−x^2}\)

12) \(\displaystyle \int\frac{dx}{a^2−x^2}\)

Answer

\(\dfrac{1}{a}\tanh^{−1}\left(\dfrac{x}{a}\right)+C\)

13) \(\displaystyle \int\frac{dx}{\sqrt{x^2+1}}\)

14) \(\displaystyle \int\frac{x \, dx}{\sqrt{x^2+1}}\)

Answer

\(\sqrt{x^2+1}+C\)

15) \(\displaystyle \int−\frac{dx}{x\sqrt{1−x^2}}\)

16) \(\displaystyle \int\frac{e^x \, dx}{\sqrt{e^{2x}−1}}\)

Answer

\(\cosh^{−1}(e^x)+C\)

17) \(\displaystyle \int−\frac{2x}{x^4−1} \, dx\)

18) Why is \(u\)-substitution referred to as a change of variable?

19) If \( f=g \circ h\), when reversing the chain rule, \(\dfrac{d}{dx}(g \circ h)(x)=g′(h(x))h′(x)\), should you take \( u=g(x)\) or \(u=h(x)\)?

Answer

\(u=h(x)\)

In exercises 20 - 24, verify each identity using differentiation. Then, using the indicated \(u\)-substitution, identify \(f\) such that the integral takes the form \(\displaystyle\int f(u)\,du\).

20) \(\displaystyle \int x\sqrt{x+1}\,dx=\frac{2}{15}(x+1)^{3/2}(3x−2)+C;\quad u=x+1\)

21) \(\displaystyle\int\frac{x^2}{\sqrt{x−1}}\,dx=\frac{2}{15}\sqrt{x−1}(3x^2+4x+8)+C,\quad (x>1);\quad u=x−1\)

Answer

\( f(u)=\dfrac{(u+1)^2}{\sqrt{u}}\)

22) \(\displaystyle\int x\sqrt{4x^2+9}\,dx=\frac{1}{12}(4x^2+9)^{3/2}+C;\quad u=4x^2+9\)

23) \(\displaystyle\int\frac{x}{\sqrt{4x^2+9}}\,dx=\frac{1}{4}\sqrt{4x^2+9}+C;\quad u=4x^2+9\)

Answer

\( du=8x\,dx;\quad f(u)=\frac{1}{8\sqrt{u}}\)

24) \(\displaystyle\int\frac{x}{(4x^2+9)^2}\,dx=−\frac{1}{8(4x^2+9)} + C;\quad u=4x^2+9\)

In exercises 25 - 34, find the antiderivative using the indicated substitution.

25) \(\displaystyle\int(x+1)^4\,dx;\quad u=x+1\)

Answer

\(\displaystyle\int(x+1)^4\,dx = \frac{1}{5}(x+1)^5+C\)

26) \(\displaystyle\int(x−1)^5\,dx;\quad u=x−1\)

27) \(\displaystyle\int(2x−3)^{−7}\,dx;\quad u=2x−3\)

Answer

\(\displaystyle\int(2x−3)^{−7}\,dx = −\frac{1}{12(2x−3)^6}+C\)

28) \(\displaystyle\int(3x−2)^{−11}\,dx;\quad u=3x−2\)

29) \(\displaystyle\int\frac{x}{\sqrt{x^2+1}}\,dx;\quad u=x^2+1\)

Answer

\(\displaystyle\int\frac{x}{\sqrt{x^2+1}}\,dx = \sqrt{x^2+1}+C\)

30) \(\displaystyle\int\frac{x}{\sqrt{1−x^2}}\,dx;\quad u=1−x^2\)

31) \(\displaystyle\int(x−1)(x^2−2x)^3\,dx;\quad u=x^2−2x\)

Answer

\(\displaystyle\int(x−1)(x^2−2x)^3\,dx = \frac{1}{8}(x^2−2x)^4+C\)

32) \(\displaystyle\int(x^2−2x)(x^3−3x^2)^2\,dx;\quad u=x^3=3x^2\)

33) \(\displaystyle\int\cos^3 \theta \,d \theta ;\quad u=\sin \theta \) (Hint: \(\cos^2 \theta =1−\sin^2 \theta \))

Answer

\(\displaystyle\int\cos^3 \theta \,d \theta = \sin \theta −\dfrac{\sin^3 \theta }{3}+C\)

34) \(\displaystyle \int\sin^3 \theta \,d \theta ;\quad u=\cos \theta \) (Hint: \(\sin^2 \theta =1−\cos^2 \theta \))

In exercises 35 - 51, use a suitable change of variables to determine the indefinite integral.

35) \(\displaystyle\int x(1−x)^{99}\,dx\)

Answer

\(\begin{align*} \displaystyle\int x(1−x)^{99}\,dx &= \frac{(1−x)^{101}}{101}−\frac{(1−x)^{100}}{100}+C \\[4pt]
&=-\frac{(1-x)^{100}}{10100}\big[ 100x + 1 \big]+C \end{align*}\)

36) \(\displaystyle\int t(1−t^2)^{10} \, dt\)

37) \(\displaystyle\int(11x−7)^{−3}\,dx\)

Answer

\(\displaystyle\int(11x−7)^{−3}\,dx = −\frac{1}{22(11x−7)^2}+C\)

38) \(\displaystyle\int(7x−11)^4\,dx\)

39) \(\displaystyle\int\cos^3 \theta \sin \theta \,d \theta \)

Answer

\(\displaystyle\int\cos^3 \theta \sin \theta \,d \theta = −\frac{\cos^4 \theta }{4}+C\)

40) \(\displaystyle\int\sin^7 \theta \cos \theta \,d \theta \)

41) \(\displaystyle\int\cos^2( \pi t)\sin( \pi t)\,dt\)

Answer

\(\displaystyle\int\cos^2( \pi t)\sin( \pi t)\,dt = −\frac{cos^3( \pi t)}{3 \pi }+C\)

42) \(\displaystyle\int\sin^2 x\cos^3 x\,dx\) (Hint: \(\sin^2 x+\cos^2 x=1\))

43) \(\displaystyle\int t\sin(t^2)\cos(t^2)\,dt\)

Answer

\(\displaystyle\int t\sin(t^2)\cos(t^2)\,dt = −\frac{1}{4}\cos^2(t^2)+C\)

44) \(\displaystyle\int t^2\cos^2(t^3)\sin(t^3)\,dt\)

45) \(\displaystyle\int\frac{x^2}{(x^3−3)^2}\,dx\)

Answer

\(\displaystyle\int\frac{x^2}{(x^3−3)^2}\,dx = −\frac{1}{3(x^3−3)}+C\)

46) \(\displaystyle\int\frac{x^3}{\sqrt{1−x^2}}\,dx\)

47) \(\displaystyle\int\frac{y^5}{(1−y^3)^{3/2}}\,dy\)

Answer

\(\displaystyle\int\frac{y^5}{(1−y^3)^{3/2}}\,dy = −\frac{2(y^3−2)}{3\sqrt{1−y^3}}+C\)

48) \(\displaystyle\int\cos \theta (1−\cos \theta )^{99}\sin \theta \,d \theta \)

49) \(\displaystyle\int(1−\cos^3 \theta )^{10}\cos^2 \theta \sin \theta \,d \theta \)

Answer

\(\displaystyle\int(1−\cos^3 \theta )^{10}\cos^2 \theta \sin \theta \,d \theta = \frac{1}{33}(1−\cos^3 \theta )^{11}+C\)

50) \(\displaystyle\int(\cos \theta −1)(\cos^2 \theta −2\cos \theta )^3\sin \theta \,d \theta \)

51) \(\displaystyle\int(\sin^2 \theta −2\sin \theta )(\sin^3 \theta −3\sin^2 \theta )^3\cos \theta \,d \theta \)

Answer

\(\displaystyle\int(\sin^2 \theta −2\sin \theta )(\sin^3 \theta −3\sin^2 \theta )^3\cos \theta \,d \theta = \frac{1}{12}(\sin^3 \theta −3\sin^2 \theta )^4+C\)

In exercises 52 - 55, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.

52) [Technology Required] \(y=3(1−x)^2\) over \([0,2]\)

53) [Technology Required] \(y=x(1−x^2)^3\) over \([−1,2]\)

Answer

\(L_{50}=−8.5779\). The exact area is \(\frac{−81}{8}\) units\(^2\).

54) [Technology Required] \(y=\sin x(1−\cos x)^2\) over \([0, \pi ]\)

55) [Technology Required] \(y=\dfrac{x}{(x^2+1)^2}\) over \([−1,1]\)

Answer

\(L_{50}=−0.006399\). The exact area is 0.

In exercises 56 - 61, use a change of variables to evaluate the definite integral.

56) \(\displaystyle\int^1_0x\sqrt{1−x^2}\,dx\)

57) \(\displaystyle\int^1_0\frac{x}{\sqrt{1+x^2}}\,dx\)

Answer

\(\displaystyle u=1+x^2,\quad du=2x\,dx,\quad \int^1_0\frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}\int^2_1u^{−1/2}du=\sqrt{2}−1\)

58) \(\displaystyle\int^2_0\frac{t}{\sqrt{5+t^2}}\,dt\)

59) \(\displaystyle\int^1_0\frac{t^2}{\sqrt{1+t^3}}\,dt\)

Answer

\(\displaystyle u=1+t^3,\quad du=3t^2,\quad \int^1_0\frac{t^2}{\sqrt{1+t^3}}\,dt = \frac{1}{3}\int^2_1u^{−1/2}du=\frac{2}{3}(\sqrt{2}−1)\)

60) \(\displaystyle\int^{ \pi /4}_0\sec^2 \theta \tan \theta \,d \theta \)

61) \(\displaystyle\int^{ \pi /4}_0\frac{\sin \theta }{\cos^4 \theta }\,d \theta \)

Answer

\(\displaystyle u=\cos \theta ,\quad du=−\sin \theta \,d \theta ,\quad \int^{ \pi /4}_0\frac{\sin \theta }{\cos^4 \theta }\,d \theta = -\int_1^{\sqrt{2}/2}u^{−4}\,du = \int^1_{\sqrt{2}/2}u^{−4}\,du=\frac{1}{3}(2\sqrt{2}−1)\)

In exercises 62 - 67, evaluate the indefinite integral \(\displaystyle \int f(x)\,dx\) with constant \(C=0\) using \(u\)-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of \(C\) that would need to be added to the antiderivative to make it equal to the definite integral \(\displaystyle F(x)=\int^x_af(t)\,dt\), with a the left endpoint of the given interval.

62) [Technology Required] \(\displaystyle\int(2x+1)e^{x^2+x−6}\,dx\) over \([−3,2]\)

63) [Technology Required] \(\displaystyle\int\frac{\cos(\ln(2x))}{x}\,dx\) on \([0,2]\)

Answer

The antiderivative is \(y=\sin(\ln(2x))\). Since the antiderivative is not continuous at \(x=0\), one cannot find a value of C that would make \(y=\sin(\ln(2x))−C\) work as a definite integral.

64) [Technology Required] \(\displaystyle \int\frac{3x^2+2x+1}{\sqrt{x^3+x^2+x+4}}\,dx\) over \([−1,2]\)

65) [Technology Required] \(\displaystyle \int\frac{\sin x}{\cos^3x}\,dx\) over \(\left[−\frac{ \pi }{3},\frac{ \pi }{3}\right]\)

Answer

The antiderivative is \(y=\frac{1}{2}\sec^2 x\). You should take \(C=−2\) so that \(F(−\frac{ \pi }{3})=0\).

66) [Technology Required] \(\displaystyle \int(x+2)e^{−x^2−4x+3}\,dx\) over \([−5,1]\)

67) [Technology Required] \(\displaystyle \int3x^2\sqrt{2x^3+1}\,dx\) over \([0,1]\)

Answer

The antiderivative is \( y=\frac{1}{3}(2x^3+1)^{3/2}\). One should take \(C=−\frac{1}{3}\).

68) If \(h(a)=h(b)\) in \(\displaystyle \int^b_ag'(h(x))h(x)\,dx\), what can you say about the value of the integral?

69) Is the substitution \(u=1−x^2\) in the definite integral \(\displaystyle \int^2_0\frac{x}{1−x^2}\,dx\) okay? If not, why not?

Answer

No, because the integrand is discontinuous at \(x=1\).

In exercises 70 - 76, use a change of variables to show that each definite integral is equal to zero.

70) \(\displaystyle \int^ \pi _0\cos^2(2 \theta )\sin(2 \theta )\,d \theta \)

71) \(\displaystyle \int^\sqrt{ \pi }_0t\cos(t^2)\sin(t^2)\,dt\)

Answer

\(u=\sin(t^2);\) the integral becomes \(\displaystyle \frac{1}{2}\int^0_0u\,du\).

72) \(\displaystyle \int^1_0(1−2t)\,dt\)

73) \(\displaystyle \int^1_0\frac{1−2t}{1+(t−\frac{1}{2})^2}\,dt\)

Answer

\(u=1+(t−\frac{1}{2})^2;\) the integral becomes \(\displaystyle −\int^{5/4}_{5/4}\frac{1}{u}\,du\).

74) \(\displaystyle \int^ \pi _0\sin\left(\left(t−\tfrac{ \pi }{2}\right)^3\right)\cos\left(t−\tfrac{ \pi }{2}\right)\,dt\)

75) \(\displaystyle \int^2_0(1−t)\cos( \pi t)\,dt\)

Answer

\(u=1−t;\) Since the integrand is odd, the integral becomes
\[\int^{−1}_1u\cos\big( \pi (1−u)\big)\,du=\int^{−1}_1u[\cos \pi \cos u−\sin \pi \sin u]\,du=−\int^{−1}_1u\cos u\,du=\int_{-1}^1u\cos u\,du=0\nonumber \]

76) \(\displaystyle \int^{3 \pi /4}_{ \pi /4}\sin^2 t\cos t\,dt\)

77) Show that the average value of \(f(x)\) over an interval \([a,b]\) is the same as the average value of \(f(cx)\) over the interval \(\left[\frac{a}{c},\frac{b}{c}\right]\) for \(c>0\).

Answer

Setting \(u=cx\) and \(du=c\,dx\) gets you \(\displaystyle \frac{1}{\frac{b}{c}−\frac{a}{c}}\int^{b/c}_{a/c}f(cx)\,dx=\frac{c}{b−a}\int^{u=b}_{u=a}f(u)\frac{du}{c}=\frac{1}{b−a}\int^b_af(u)\,du\).

78) Find the area under the graph of \(f(t)=\dfrac{t}{(1+t^2)^a}\) between \(t=0\) and \(t=x\) where \(a>0\) and \(a \neq 1\) is fixed, and evaluate the limit as \(x \to \infty \).

79) Find the area under the graph of \(g(t)=\dfrac{t}{(1−t^2)^a}\) between \(t=0\) and \(t=x\), where \(0<x<1\) and \(a>0\) is fixed. Evaluate the limit as \(x \to 1\).

Answer

\(\displaystyle \int^x_0g(t)\,dt=\frac{1}{2}\int^1_{u=1−x^2} \frac{du}{u^a}=\frac{1}{2(1−a)}u^{1−a}∣1u=\frac{1}{2(1−a)}(1−(1−x^2)^{1−a})\) As \(x \to 1\) the limit is \(\dfrac{1}{2(1−a)}\) if \(a<1\), and the limit diverges to \(+ \infty \) if \(a>1\).

80) The area of a semicircle of radius \(1\) can be expressed as \(\displaystyle \int^1_{−1}\sqrt{1−x^2}\,dx\). Use the substitution \(x=\cos t\) to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.

81) The area of the top half of an ellipse with a major axis that is the \(x\)-axis from \(x=−1\) to a and with a minor axis that is the \(y\)-axis from \(y=−b\) to \(y=b\) can be written as \(\displaystyle \int^a_{−a}b\sqrt{1−\frac{x^2}{a^2}}\,dx\). Use the substitution \(x=a\cos t\) to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

Answer

\(\displaystyle \int^{t=0}_{t= \pi }b\sqrt{1−\cos^2 t} \times (−a\sin t)\,dt=\int^{t= \pi }_{t=0}ab\sin^2 t\,dt\)

82) [Technology Required] The following graph is of a function of the form \( f(t)=a\sin(nt)+b\sin(mt)\). Estimate the coefficients \(a\) and \(b\) and the frequency parameters \(n\) and \(m\). Use these estimates to approximate \(\displaystyle \int^ \pi _0f(t)\,dt\).

83) [Technology Required] The following graph is of a function of the form \(f(x)=a\cos(nt)+b\cos(mt)\). Estimate the coefficients \(a\) and \(b\) and the frequency parameters \(n\) and \(m\). Use these estimates to approximate \(\displaystyle \int^ \pi _0f(t)\,dt\).

Answer

\(f(t)=2\cos(3t)−\cos(2t);\quad \displaystyle \int^{ \pi /2}_0(2\cos(3t)−\cos(2t))\,dt=−\frac{2}{3}\)