1.5E: Exercises for Section 1.5

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

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

Answer: $u=h(x)$

In exercises 3 - 7, verify each identity using differentiation. Then, using the indicated $u$-substitution, identify $f$ such that the integral takes the form $\displaystyle∫f(u)\,du.$

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

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

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

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

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

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

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

In exercises 8 - 17, find the antiderivative using the indicated substitution.

8) $\displaystyle∫(x+1)^4\,dx;\quad u=x+1$

Answer: $\displaystyle∫(x+1)^4\,dx = \frac{1}{5}(x+1)^5+C$

9) $\displaystyle∫(x−1)^5\,dx;\quad u=x−1$

10) $\displaystyle∫(2x−3)^{−7}\,dx;\quad u=2x−3$

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

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

12) $\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx;\quad u=x^2+1$

Answer: $\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx = \sqrt{x^2+1}+C$

13) $\displaystyle∫\frac{x}{\sqrt{1−x^2}}\,dx;\quad u=1−x^2$

14) $\displaystyle∫(x−1)(x^2−2x)^3\,dx;\quad u=x^2−2x$

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

15) $\displaystyle∫(x^2−2x)(x^3−3x^2)^2\,dx;\quad u=x^3=3x^2$

16) $\displaystyle∫\cos^3 θ\,dθ;\quad u=\sin θ$ (Hint: $\cos^2 θ=1−\sin^2 θ$)

Answer: $\displaystyle∫\cos^3 θ\,dθ = \sin θ−\dfrac{\sin^3 θ}{3}+C$

17) $\displaystyle ∫\sin^3 θ\,dθ;\quad u=\cos θ$ (Hint: $\sin^2 θ=1−\cos^2θ$)

In exercises 18 - 34, use a suitable change of variables to determine the indefinite integral.

18) $\displaystyle∫x(1−x)^{99}\,dx$

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

19) $\displaystyle∫t(1−t^2)^{10}dt$

20) $\displaystyle∫(11x−7)^{−3}\,dx$

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

21) $\displaystyle∫(7x−11)^4\,dx$

22) $\displaystyle∫\cos^3 θ\sin θ\,dθ$

Answer: $\displaystyle∫\cos^3 θ\sin θ\,dθ = −\frac{\cos^4 θ}{4}+C$

23) $\displaystyle∫\sin^7 θ\cos θ\,dθ$

24) $\displaystyle∫\cos^2(πt)\sin(πt)\,dt$

Answer: $\displaystyle∫\cos^2(πt)\sin(πt)\,dt = −\frac{cos^3(πt)}{3π}+C$

25) $\displaystyle∫\sin^2 x\cos^3 x\,dx$ (Hint: $\sin^2 x+\cos^2 x=1$)

26) $\displaystyle∫t\sin(t^2)\cos(t^2)\,dt$

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

27) $\displaystyle∫t^2\cos^2(t^3)\sin(t^3)\,dt$

28) $\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx$

Answer: $\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx = −\frac{1}{3(x^3−3)}+C$

29) $\displaystyle∫\frac{x^3}{\sqrt{1−x^2}}\,dx$

30) $\displaystyle∫\frac{y^5}{(1−y^3)^{3/2}}\,dy$

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

31) $\displaystyle∫\cos θ(1−\cos θ)^{99}\sin θ\,dθ$

32) $\displaystyle∫(1−\cos^3 θ)^{10}\cos^2 θ\sin θ\,dθ$

Answer: $\displaystyle∫(1−\cos^3 θ)^{10}\cos^2 θ\sin θ\,dθ = \frac{1}{33}(1−\cos^3 θ)^{11}+C$

33) $\displaystyle∫(\cos θ−1)(\cos^2 θ−2\cos θ)^3\sin θ\,dθ$

34) $\displaystyle∫(\sin^2 θ−2\sin θ)(\sin^3 θ−3\sin^2 θ)^3\cos θ\,dθ$

Answer: $\displaystyle∫(\sin^2 θ−2\sin θ)(\sin^3 θ−3\sin^2 θ)^3\cos θ\,dθ = \frac{1}{12}(\sin^3 θ−3\sin^2 θ)^4+C$

In exercises 35 - 38, 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 $[0,2]$

36) [T] $y=x(1−x^2)^3$ over $[−1,2]$

Answer: $L_{50}=−8.5779.$ The exact area is $\frac{−81}{8}$ units$^2$.

37) [T] $y=\sin x(1−\cos x)^2$ over $[0,π]$

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

Answer: $L_{50}=−0.006399$. The exact area is 0.

In exercises 39 - 44, use a change of variables to evaluate the definite integral.

39) $\displaystyle∫^1_0x\sqrt{1−x^2}\,dx$

40) $\displaystyle∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx$

Answer: $\displaystyle u=1+x^2,\quad du=2x\,dx,\quad ∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1$

41) $\displaystyle∫^2_0\frac{t}{\sqrt{5+t^2}}\,dt$

42) $\displaystyle∫^1_0\frac{t^2}{\sqrt{1+t^3}}\,dt$

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

43) $\displaystyle∫^{π/4}_0\sec^2 θ\tan θ\,dθ$

44) $\displaystyle∫^{π/4}_0\frac{\sin θ}{\cos^4 θ}\,dθ$

Answer: $\displaystyle u=\cos θ,\quad du=−\sin θ\,dθ,\quad \int^{π/4}_0\frac{\sin θ}{\cos^4 θ}\,dθ = -∫_1^{\sqrt{2}/2}u^{−4}\,du = ∫^1_{\sqrt{2}/2}u^{−4}\,du=\frac{1}{3}(2\sqrt{2}−1)$

In exercises 45 - 50, evaluate the indefinite integral $\displaystyle ∫f(x)\,dx$ 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 $\displaystyle F(x)=∫^x_af(t)\,dt$, with a the left endpoint of the given interval.

45) [T] $\displaystyle∫(2x+1)e^{x^2+x−6}\,dx$ over $[−3,2]$

46) [T] $\displaystyle∫\frac{\cos(\ln(2x))}{x}\,dx$ on $[0,2]$

Answer

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

The antiderivative is $y=\sin(\ln(2x))$. 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.

47) [T] $\displaystyle ∫\frac{3x^2+2x+1}{\sqrt{x^3+x^2+x+4}}\,dx$ over $[−1,2]$

48) [T] $\displaystyle ∫\frac{\sin x}{\cos^3x}\,dx$ over $\left[−\frac{π}{3},\frac{π}{3}\right]$

Answer

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

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

49) [T] $\displaystyle ∫(x+2)e^{−x^2−4x+3}\,dx$ over $[−5,1]$

50) [T] $\displaystyle ∫3x^2\sqrt{2x^3+1}\,dx$ over $[0,1]$

Answer

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

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

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

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

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

In exercises 53 - 59, use a change of variables to show that each definite integral is equal to zero.

53) $\displaystyle ∫^π_0\cos^2(2θ)\sin(2θ)\,dθ$

54) $\displaystyle ∫^\sqrt{π}_0t\cos(t^2)\sin(t^2)\,dt$

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

55) $\displaystyle ∫^1_0(1−2t)\,dt$

56) $\displaystyle ∫^1_0\frac{1−2t}{1+(t−\frac{1}{2})^2}\,dt$

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

57) $\displaystyle ∫^π_0\sin\left(\left(t−\tfrac{π}{2}\right)^3\right)\cos\left(t−\tfrac{π}{2}\right)\,dt$

58) $\displaystyle ∫^2_0(1−t)\cos(πt)\,dt$

Answer: $u=1−t;$ Since the integrand is odd, the integral becomes
\[∫^{−1}_1u\cos\big(π(1−u)\big)\,du=∫^{−1}_1u[\cos π\cos u−\sin π\sin u]\,du=−∫^{−1}_1u\cos u\,du=∫_{-1}^1u\cos u\,du=0\nonumber \]

59) $\displaystyle ∫^{3π/4}_{π/4}\sin^2 t\cos t\,dt$

60) 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 $\displaystyle \frac{1}{\frac{b}{c}−\frac{a}{c}}∫^{b/c}_{a/c}f(cx)\,dx=\frac{c}{b−a}∫^{u=b}_{u=a}f(u)\frac{du}{c}=\frac{1}{b−a}∫^b_af(u)\,du.$

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 $a≠1$ is fixed, and evaluate the limit as $x→∞$.

62) 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 $x→1$.

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

63) The area of a semicircle of radius $1$ can be expressed as $\displaystyle ∫^1_{−1}\sqrt{1−x^2}\,dx$. 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) Use ChatGPT (or another LLM) to analyze when integration by substitution works for evaluating integrals, when it does not work, and how to choose the substitution when it does work. Submit your prompt(s), a summary of the output from ChatGPT, and your analysis of how accurate this output is based on the classwork and homework.

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