Problem Set: Substitution

1. Why is $u$ -substitution referred to as change of variable?

2. If $f=g\circ h,$ when reversing the chain rule, $\frac{d}{dx}(g\circ h)(x)={g}^{\prime }(h(x)){h}^{\prime }(x),$ should you take $u=g(x)$ or $u=h(x)?$

Show Solution

$u=h(x)$

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

3. $\displaystyle\int x\sqrt{x+1}dx=\frac{2}{15}{(x+1)}^{3\text{/}2}(3x-2)+C; \,\,\, u=x+1$

4. For $x>1$ : $\displaystyle\int \frac{{x}^{2}}{\sqrt{x-1}}dx = \frac{2}{15}\sqrt{x-1}(3{x}^{2}+4x+8)+C; \,\,\, u=x-1$

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

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

6. $\displaystyle\int \frac{x}{\sqrt{4{x}^{2}+9}}dx=\frac{1}{4}\sqrt{4{x}^{2}+9}+C;\,\,\, u=4{x}^{2}+9$

Show Solution

$du=8xdx;f(u)=\frac{1}{8\sqrt{u}}$

7. $\displaystyle\int \frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\frac{1}{8(4{x}^{2}+9)};\,\,\, u=4{x}^{2}+9$

In the following exercises, find the antiderivative using the indicated substitution.

8. $\displaystyle\int {(x+1)}^{4}dx; \,\,\, u=x+1$

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$\frac{1}{5}{(x+1)}^{5}+C$

9. $\displaystyle\int {(x-1)}^{5}dx; \,\,\, u=x-1$

10. $\displaystyle\int {(2x-3)}^{-7}dx; \,\,\, u=2x-3$

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$-\frac{1}{12{(3-2x)}^{6}}+C$

11. $\displaystyle\int {(3x-2)}^{-11}dx; \,\,\, u=3x-2$

12. $\displaystyle\int \frac{x}{\sqrt{{x}^{2}+1}}dx; \,\,\, u={x}^{2}+1$

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$\sqrt{{x}^{2}+1}+C$

13. $\displaystyle\int \frac{x}{\sqrt{1-{x}^{2}}}dx; \,\,\, u=1-{x}^{2}$

14. $\displaystyle\int (x-1){({x}^{2}-2x)}^{3}dx; \,\,\, u={x}^{2}-2x$

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$\frac{1}{8}{({x}^{2}-2x)}^{4}+C$

15. $\displaystyle\int ({x}^{2}-2x){({x}^{3}-3{x}^{2})}^{2}dx; \,\,\, u={x}^{3}-3{x}^{2}$

16. $\displaystyle\int { \cos }^{3}\theta d\theta ; \,\,\, u= \sin \theta$

Hint

( $\cos ^{2}\theta =1-{ \sin }^{2}\theta$ )

Show Solution

$\sin \theta -\frac{{ \sin }^{3}\theta }{3}+C$

17. $\displaystyle\int { \sin }^{3}\theta d\theta ; \,\,\, u= \cos \theta$

Hint

$\sin ^{2}\theta =1-{ \cos }^{2}\theta$

In the following exercises, use a suitable change of variables to determine the indefinite integral.

18. $\displaystyle\int x{(1-x)}^{99}dx$

Show Solution

$\frac{{(1-x)}^{101}}{101}-\frac{{(1-x)}^{100}}{100}+C$

19. $\displaystyle\int t{(1-{t}^{2})}^{10}dt$

20. $\displaystyle\int {(11x-7)}^{-3}dx$

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$-\frac{1}{22(7-11{x}^{2})}+C$

21. $\displaystyle\int {(7x-11)}^{4}dx$

22. $\displaystyle\int { \cos }^{3}\theta \sin \theta d\theta$

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$-\frac{{ \cos }^{4}\theta }{4}+C$

23. $\displaystyle\int { \sin }^{7}\theta \cos \theta d\theta$

24. $\displaystyle\int { \cos }^{2}(\pi t) \sin (\pi t)dt$

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$-\frac{{ \cos }^{3}(\pi t)}{3\pi }+C$

25. $\displaystyle\int { \sin }^{2}x{ \cos }^{3}xdx$

Hint

$\sin ^{2}x+{ \cos }^{2}x=1$

26. $\displaystyle\int t \sin ({t}^{2}) \cos ({t}^{2})dt$

Show Solution

$-\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{ \cos }^{2}({t}^{2})+C$

27. $\displaystyle\int {t}^{2}{ \cos }^{2}({t}^{3}) \sin ({t}^{3})dt$

28. $\displaystyle\int \frac{{x}^{2}}{{({x}^{3}-3)}^{2}}dx$

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

29. $\displaystyle\int \frac{{x}^{3}}{\sqrt{1-{x}^{2}}}dx$

30. $\displaystyle\int \frac{{y}^{5}}{{(1-{y}^{3})}^{3\text{/}2}}dy$

Show Solution

$-\frac{2({y}^{3}-2)}{3\sqrt{1-{y}^{3}}}$

31. ${\displaystyle\int \cos \theta (1- \cos \theta )}^{99} \sin \theta d\theta$

32. ${\displaystyle\int (1-{ \cos }^{3}\theta )}^{10}{ \cos }^{2}\theta \sin \theta d\theta$

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$\frac{1}{33}{(1-{ \cos }^{3}\theta )}^{11}+C$

33. $\displaystyle\int ( \cos \theta -1){({ \cos }^{2}\theta -2 \cos \theta )}^{3} \sin \theta d\theta$

34. $\displaystyle\int ({ \sin }^{2}\theta -2 \sin \theta ){({ \sin }^{3}\theta -3{ \sin }^{2}\theta )}^{3} \cos \theta d\theta$

Show Solution

$\frac{1}{12}{({ \sin }^{3}\theta -3{ \sin }^{2}\theta )}^{4}+C$

In the following exercises, 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.

35. [T] $y=3{(1-x)}^{2}$ over $\left[0,2\right]$

36. [T] $y=x{(1-{x}^{2})}^{3}$ over $\left[-1,2\right]$

Show Solution

${L}_{50}=-8.5779.$ The exact area is $\frac{-81}{8}$

37. [T] $y= \sin x{(1- \cos x)}^{2}$ over $\left[0,\pi \right]$

38. [T] $y=\dfrac{x}{{({x}^{2}+1)}^{2}}$ over $\left[-1,1\right]$

Show Solution

${L}_{50}=-0.006399$ … The exact area is 0.

In the following exercises, use a change of variables to evaluate the definite integral.

39. ${\displaystyle\int }_{0}^{1}x\sqrt{1-{x}^{2}}dx$

40. ${\displaystyle\int }_{0}^{1}\dfrac{x}{\sqrt{1+{x}^{2}}}dx$

Show Solution

$u=1+{x}^{2},du=2xdx,\frac{1}{2}{\displaystyle\int }_{1}^{2}{u}^{-1\text{/}2}du=\sqrt{2}-1$

41. ${\displaystyle\int }_{0}^{2}\dfrac{t^2}{\sqrt{5+{t}^{2}}}dt$

42. ${\displaystyle\int }_{0}^{1}\dfrac{t^2}{\sqrt{1+{t}^{3}}}dt$

Show Solution

$u=1+{t}^{3},du=3{t}^{2}du,\frac{1}{3}{\displaystyle\int }_{1}^{2}{u}^{-1\text{/}2}du=\frac{2}{3}(\sqrt{2}-1)$

43. ${\displaystyle\int }_{0}^{\pi \text{/}4}{ \sec }^{2}\theta \tan \theta d\theta$

44. ${\displaystyle\int }_{0}^{\pi \text{/}4}\dfrac{ \sin \theta }{{ \cos }^{4}\theta }d\theta$

Show Solution

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

In the following exercises, 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 $F(x)={\displaystyle\int }_{a}^{x}f(t)dt,$ with $a$ the left endpoint of the given interval.

45. [T] $\displaystyle\int (2x+1){e}^{{x}^{2}+x-6}dx$ over $\left[-3,2\right]$

46. [T] $\displaystyle\int \frac{ \cos (\text{ln}(2x))}{x}dx$ on $\left[0,2\right]$

Show Solution

Two graphs. The first shows the function f(x) = cos(ln(2x)) / x, which increases sharply over the approximate interval (0,.25) and then decreases gradually to the x axis. The second shows the function f(x) = sin(ln(2x)), which decreases sharply on the approximate interval (0, .25) and then increases in a gently curve into the first quadrant.

The antiderivative is

$y= \sin (\text{ln}(2x)).$ Since the antiderivative is not continuous at

$x=0,$ one cannot find a value of C that would make

$y= \sin (\text{ln}(2x))-C$ work as a definite integral.

47. [T] $\displaystyle\int \frac{3{x}^{2}+2x+1}{\sqrt{{x}^{3}+{x}^{2}+x+4}}dx$ over $\left[-1,2\right]$

48. [T] $\displaystyle\int \frac{ \sin x}{{ \cos }^{3}x}dx$ over $\left[-\frac{\pi }{3},\frac{\pi }{3}\right]$

Show Solution

Two graphs. The first is the function f(x) = sin(x) / cos(x)^3 over [-5pi/16, 5pi/16]. It is an increasing concave down function for values less than zero and an increasing concave up function for values greater than zero. The second is the fuction f(x) = ½ sec(x)^2 over the same interval. It is a wide, concave up curve which decreases for values less than zero and increases for values greater than zero.

The antiderivative is

$y=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{ \sec }^{2}x.$ You should take

$C=-2$ so that

$F(-\frac{\pi }{3})=0.$

49. [T] $\displaystyle\int (x+2){e}^{\text{−}{x}^{2}-4x+3}dx$ over $\left[-5,1\right]$

50. [T] $\displaystyle\int 3{x}^{2}\sqrt{2{x}^{3}+1}dx$ over $\left[0,1\right]$

Show Solution

Two graphs. The first shows the function f(x) = 3x^2 * sqrt(2x^3 + 1). It is an increasing concave up curve starting at the origin. The second shows the function f(x) = 1/3 * (2x^3 + 1)^(1/3). It is an increasing concave up curve starting at about 0.3.

The antiderivative is

$y=\frac{1}{3}{(2{x}^{3}+1)}^{3\text{/}2}.$ One should take

$C=-\frac{1}{3}.$

51. If $h(a)=h(b)$ in ${\displaystyle\int }_{a}^{b}g\text{‘}(h(x))h(x)dx,$ what can you say about the value of the integral?

52. Is the substitution $u=1-{x}^{2}$ in the definite integral ${\displaystyle\int }_{0}^{2}\dfrac{x}{1-{x}^{2}}dx$ okay? If not, why not?

Show Solution

No, because the integrand is discontinuous at $x=1.$

In the following exercises, use a change of variables to show that each definite integral is equal to zero.

53. ${\displaystyle\int }_{0}^{\pi }{ \cos }^{2}(2\theta ) \sin (2\theta )d\theta$

54. ${\displaystyle\int }_{0}^{\sqrt{\pi }}t \cos ({t}^{2}) \sin ({t}^{2})dt$

Show Solution

$u= \sin ({t}^{2});$ the integral becomes $\frac{1}{2}{\displaystyle\int }_{0}^{0}udu.$

55. ${\displaystyle\int }_{0}^{1}(1-2t)dt$

56. ${\displaystyle\int }_{0}^{1}\dfrac{1-2t}{(1+{(t-\frac{1}{2})}^{2})}dt$

Show Solution

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

57. ${\displaystyle\int }_{0}^{\pi } \sin ({(t-\frac{\pi }{2})}^{3}) \cos (t-\frac{\pi }{2})dt$

58. ${\displaystyle\int }_{0}^{2}(1-t) \cos (\pi t)dt$

Show Solution

$u=1-t;$ the integral becomes

$\begin{array}{l}{\displaystyle\int }_{1}^{-1}u \cos (\pi (1-u))du\hfill \\ ={\displaystyle\int }_{1}^{-1}u\left[ \cos \pi \cos u- \sin \pi \sin u\right]du\hfill \\ =\text{−}{\displaystyle\int }_{1}^{-1}u \cos udu\hfill \\ ={\displaystyle\int }_{-1}^{1}u \cos udu=0\hfill \end{array}$
since the integrand is odd.

59. ${\displaystyle\int }_{\pi \text{/}4}^{3\pi \text{/}4}{ \sin }^{2}t \cos tdt$

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

Show Solution

Setting $u=cx$ and $du=cdx$ gets you $\frac{1}{\frac{b}{c}-\frac{a}{c}}{\displaystyle\int }_{a\text{/}c}^{b\text{/}c}f(cx)dx=\frac{c}{b-a}{\displaystyle\int }_{u=a}^{u=b}f(u)\frac{du}{c}=\frac{1}{b-a}{\displaystyle\int }_{a}^{b}f(u)du.$

61. 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 $a\ne 1$ is fixed, and evaluate the limit as $x\to \infty .$

62. Find the area under the graph of $g(t)=\dfrac{t}{{(1-{t}^{2})}^{a}}$ between $t=0$ and $t=x,$ where [latex]00[/latex] is fixed. Evaluate the limit as $x\to 1.$

Show Solution

${\displaystyle\int }_{0}^{x}g(t)dt=\frac{1}{2}{\displaystyle\int }_{u=1-{x}^{2}}^{1}\frac{du}{{u}^{a}}=\frac{1}{2(1-a)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\frac{1}{2(1-a)}(1-{(1-{x}^{2})}^{1-a}).$ As $x\to 1$ the limit is $\frac{1}{2(1-a)}$ if $a<1,$ and the limit diverges to +∞ if $a>1.$

63. 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.

64. 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 $y=\text{−}b$ to $b$ can be written as ${\displaystyle\int }_{\text{−}a}^{a}b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}dx.$ 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.

Show Solution

${\displaystyle\int }_{t=\pi }^{0}b\sqrt{1-{ \cos }^{2}t}×(\text{−}a \sin t)dt={\displaystyle\int }_{t=0}^{\pi }ab{ \sin }^{2}tdt$

65. [T] 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 ${\displaystyle\int }_{0}^{\pi }f(t)dt.$

66. [T] 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 ${\displaystyle\int }_{0}^{\pi }f(t)dt.$

Show Solution

$f(t)=2 \cos (3t)- \cos (2t);{\displaystyle\int }_{0}^{\pi \text{/}2}(2 \cos (3t)- \cos (2t))=-\frac{2}{3}$

Integration: Supplemental Content

Problem Set: Substitution

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