5.5E: Exercises for Section 5.5

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

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

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

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) 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 $\displaystyle ∫^a_{−a}b\sqrt{1−\frac{x^2}{a^2}}\,dx$. 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

$\displaystyle ∫^{t=0}_{t=π}b\sqrt{1−\cos^2 t}×(−a\sin t)\,dt=∫^{t=π}_{t=0}ab\sin^2 t\,dt$

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 $\displaystyle ∫^π_0f(t)\,dt$.

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 $\displaystyle ∫^π_0f(t)\,dt.$

Answer

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