Skip to main content

5.5E and 5.6E u-Substitution Exercises

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

5.5: Substitution

In exercises 1 - 16, find the antiderivative.

1) $$\displaystyle∫(x+1)^4\,dx$$

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

2) $$\displaystyle∫(x−1)^5\,dx$$

3) $$\displaystyle∫(2x−3)^{−7}\,dx$$

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

4) $$\displaystyle∫(3x−2)^{−11}\,dx$$

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

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

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

7) $$\displaystyle∫(x−1)(x^2−2x)^3\,dx$$

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

8) $$\displaystyle∫(x^2−2x)(x^3−3x^2)^2\,dx$$

9) $$\displaystyle\int\cos^3 θ\,dθ$$ (Hint: $$\cos^2 θ=1−\sin^2 θ$$)

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

10) $$\displaystyle\int\sin^3 θ\,dθ$$ (Hint: $$\sin^2 θ=1−\cos^2 θ$$)

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

Answer:
$$\displaystyle\int x(1−x)^{99}\,dx\quad$$ $$=\quad\displaystyle\frac{(1−x)^{101}}{101}−\frac{(1−x)^{100}}{100}+C$$

12) $$\displaystyle∫t(1−t^2)^{10}\,dt$$

13) $$\displaystyle∫(11x−7)^{−3}\,dx$$

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

14) $$\displaystyle∫(7x−11)^4\,dx$$

15) $$\displaystyle\int\cos^3 θ\sin θ\,dθ$$

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

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

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

u-Substitution with Definite Integrals

In exercises 17 - 22, evaluate the definite integral.

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

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

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

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

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

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

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

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

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

u-Substitution with a Twist

In exercises 23 - 28, find the antiderivative. Then check your answer by showing its derivative can be simplified to the original integrand.

23) $$\displaystyle ∫x\sqrt{x+1}\,dx$$

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

24) $$\displaystyle \int\frac{x}{\sqrt{3x+1}}\,dx$$

Answer:
$$\displaystyle \int\frac{x}{\sqrt{3x+1}}\,dx \quad = \quad \frac{2}{27}\sqrt{3x+1}\big(3x - 2\big) + C$$

25) $$\displaystyle \int\frac{x+2}{(x-1)^{3/2}}\,dx$$

Answer:
$$\displaystyle \int\frac{x+2}{(x-1)^{3/2}}\,dx \quad = \quad \frac{2(x - 4)}{\sqrt{x-1}} + C$$

Check (using quotient rule):
\begin{align*} \dfrac{d}{dx}\left( \frac{2(x - 4)}{\sqrt{x-1}} + C \right) \quad &= \quad \dfrac{2\sqrt{x-1} - 2(x - 4)\cdot\frac{1}{2}(x-1)^{-1/2}}{x-1} \\[5pt] &=\dfrac{(x-1)^{-1/2}\big( 2(x-1) - (x - 4) \big)}{x-1} \\[5pt] &=\dfrac{2x - 2 - x + 4}{(x-1)^{3/2}} \\[5pt] &=\dfrac{x+2}{(x-1)^{3/2}} \quad \checkmark \end{align*}

26) $$\displaystyle \int t^2\sqrt{3 - t} \,dt$$

Answer:
\displaystyle \begin{align*} \int t^2\sqrt{3 - t} \,dt \quad &= -6(3 - t)^{3/2} + \frac{12}{5}(3 - t)^{5/2} - \frac{2}{7}(3 - t)^{7/2} + C \\[5pt] &= -\frac{2}{35}(3 - t)^{3/2}\big[5t^2 + 12t + 24\big] + C \end{align*}

Check (using product rule):
\begin{align*} \dfrac{d}{dx}\left( -\frac{2}{35}(3 - t)^{3/2}\big[5t^2 + 12t + 24\big] + C \right) &=-\frac{2}{35}\bigg[ (3 - t)^{3/2}(10t + 12) - \frac{3}{2}(3 - t)^{1/2}\big[5t^2 + 12t + 24\big]\bigg] \\[5pt] &=-\frac{1}{35}(3 - t)^{1/2}\bigg[ 2(3 - t)(10t + 12) - 3\big[5t^2 + 12t + 24\big]\bigg] \\[5pt] &=-\frac{1}{35}(3 - t)^{1/2}\big[ 4(3 - t)(5t + 6) - 15t^2 - 36t - 72\big] \\[5pt] &=-\frac{1}{35}(3 - t)^{1/2}\big[ 4(18 + 9t - 5t^2) - 15t^2 - 36t - 72\big] \\[5pt] &=-\frac{1}{35}(3 - t)^{1/2}\big[ 72 + 36t - 20t^2 - 15t^2 - 36t - 72\big] \\[5pt] &=-\frac{1}{35}(3 - t)^{1/2}\big[ -35t^2\big] \\[5pt] &=t^2\sqrt{3 - t} \quad \checkmark \end{align*}

27) $$\displaystyle \int t^3\,\sqrt{t^2 + 5} \,dt$$

Answer:
\displaystyle \begin{align*} \int t^3\,\sqrt{t^2 + 5} \,dt \quad &= \frac{1}{2}\int (u - 5)u^{1/2}\, du \\[5pt] &= \frac{(t^2 + 5)^{5/2}}{5} - \frac{5}{3}(t^2 + 5)^{3/2} + C\\[5pt] &= \frac{(t^2 + 5)^{3/2}}{15}\big[ 3t^2 - 10 \big] + C \end{align*}

Check (using product rule):
\begin{align*} \dfrac{d}{dx}\left( \frac{(t^2 + 5)^{3/2}}{15}\big[ 3t^2 - 10 \big] + C \right) &=\frac{6t(t^2 + 5)^{3/2}}{15} + (3t^2 - 10)\cdot\frac{3}{2}\frac{(t^2 + 5)^{1/2}}{15}\cdot 2t \\[5pt] &=\frac{3t(t^2+5)^{1/2}}{15}\big[ 2(t^2+5) + (3t^2-10) \big] \\[5pt] &=\frac{t}{5}\sqrt{t^2 + 5} \big[ 5t^2 \big] \\[5pt] &=t^3\,\sqrt{t^2 + 5} \quad \checkmark \end{align*}

28) $$\displaystyle \int x\sqrt[3]{x - 2} \,dx$$

Answer:
\displaystyle \begin{align*} \int x\sqrt[3]{x - 2} \,dx \quad &= \frac{3}{7}(x-2)^{7/3}+\frac{3}{2}(x-2)^{4/3}+ C \\[5pt] &= \frac{3}{14}(x-2)^{4/3}[2x+3] + C \end{align*}

Check (using product rule):
\begin{align*} \dfrac{d}{dx}\left( \frac{3}{14}(x-2)^{4/3}[2x+3] + C \right) &=\frac{3}{7}(x-2)^{4/3} + (2x+3)\cdot\frac{3}{14}\cdot\frac{4}{3}(x-2)^{1/3} \\[5pt] &=\frac{3}{7}(x-2)^{4/3} + \frac{2}{7}(x-2)^{1/3}(2x+3) \\[5pt] &=\frac{1}{7}(x-2)^{1/3}\big[ 3(x-2) + 2(2x+3) \big] \\[5pt] &=\frac{1}{7}(x-2)^{1/3}\big[ 3x-6 + 4x+6 \big] \\[5pt] &=\frac{1}{7}(x-2)^{1/3}\big[ 7x \big] \\[5pt] &=x\sqrt[3]{x - 2} \quad \checkmark \end{align*}

5.6: Integrals Involving Exponential and Logarithmic Functions

For exercises 1 - 8, compute each indefinite integral.

1) $$\displaystyle ∫e^{2x}\,dx$$

2) $$\displaystyle ∫e^{−3x}\,dx$$

Answer:
$$\displaystyle ∫e^{−3x}\,dx \quad = \quad \frac{−1}{3}e^{−3x}+C$$

3) $$\displaystyle ∫2^x\,dx$$

4) $$\displaystyle ∫3^{−x}\,dx$$

Answer:
$$\displaystyle ∫3^{−x}\,dx \quad = \quad −\frac{3^{−x}}{\ln 3}+C$$

5) $$\displaystyle ∫\frac{1}{2x}\,dx$$

6) $$\displaystyle ∫\frac{2}{x}\,dx$$

Answer:
$$\displaystyle ∫\frac{2}{x}\,dx \quad = \quad 2\ln x+C \quad = \quad \ln(x^2)+C$$

7) $$\displaystyle ∫\frac{1}{x^2}\,dx$$

8) $$\displaystyle ∫\frac{1}{\sqrt{x}}\,dx$$

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

In exercises 9 - 16, find each indefinite integral by using appropriate substitutions.

9) $$\displaystyle ∫\frac{\ln x}{x}\,dx$$

10) $$\displaystyle ∫\frac{dx}{x(\ln x)^2}$$

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

11) $$\displaystyle ∫xe^{−x^2}\,dx$$

12) $$\displaystyle ∫x^2e^{−x^3}\,dx$$

Answer:
$$\displaystyle ∫x^2e^{−x^3}\,dx \quad = \quad \dfrac{−e^{−x^3}}{3}+C$$

13) $$\displaystyle ∫e^{\sin x}\cos x\,dx$$

14) $$\displaystyle ∫e^{\tan x}\sec^2 x\,dx$$

Answer:
$$\displaystyle ∫e^{\tan x}\sec^2 x\,dx\quad = \quad e^{\tan x}+C$$

15) $$\displaystyle ∫\frac{e^{\ln x}}{x}\,dx$$

16) $$\displaystyle ∫\frac{e^{\ln(1−t)}}{1−t}\,dt$$

Answer:
$$\displaystyle ∫\frac{e^{\ln(1−t)}}{1−t}\,dt = \int \frac{1-t}{1-t}\,dt = \int 1\, dt \quad = \quad t+C$$

More u-Substitutions with Definite Integrals

In exercises 17 - 21, evaluate the definite integral.

17) $$\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}\,dx$$

Answer:
$$\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}\,dx \quad = \quad \frac{1}{3}\ln(\frac{26}{7})$$

18) $$\displaystyle ∫^{π/4}_0\tan x\,dx$$

19) $$\displaystyle ∫^{π/3}_0\frac{\sin x−\cos x}{\sin x+\cos x}\,dx$$

Answer:
$$\displaystyle ∫^{π/3}_0\frac{\sin x−\cos x}{\sin x+\cos x}\,dx \quad = \quad \ln(\sqrt{3}−1)$$

20) $$\displaystyle ∫^{π/2}_{π/6}\csc x\,dx$$

21) $$\displaystyle ∫^{π/3}_{π/4}\cot x\,dx$$

Answer:
$$\displaystyle ∫^{π/3}_{π/4}\cot x\,dx \quad = \quad \frac{1}{2}\ln\frac{3}{2}$$

Some Interesting u-Substitutions

In exercises 22 - 29, integrate using the indicated substitution.

22) $$\displaystyle ∫\frac{x}{x−100}\,dx;\quad u=x−100$$

23) $$\displaystyle ∫\frac{y−1}{y+1}\,dy;\quad u=y+1$$

Answer:
$$\displaystyle ∫\frac{y−1}{y+1}\,dy \quad = \quad y−2\ln|y+1|+C$$

24) $$\displaystyle ∫\frac{1−x^2}{3x−x^3}\,dx;\quad u=3x−x^3$$

25) $$\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx;\quad u=\sin x−\cos x$$

Answer:
$$\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx \quad=\quad \ln|\sin x−\cos x|+C$$

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

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

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

28) $$\displaystyle \int \frac{\sqrt{x}}{\sqrt{x} + 2}\,dx; \quad u = \sqrt{x} + 2$$

Answer:
$$\displaystyle \int \frac{\sqrt{x}}{\sqrt{x} + 2}\,dx \quad = \quad \left( \sqrt{x} + 2 \right)^2 - 8\left( \sqrt{x} + 2 \right) + 8\ln\left( \sqrt{x} + 2 \right) + C$$

29) $$\displaystyle \int e^x\sec(e^x+1)\tan(e^x+1)\,dx; \quad u = e^{x} + 1$$

Answer:
$$\displaystyle \int e^x\sec(e^x+1)\tan(e^x+1)\,dx \quad = \quad \sec(e^x+1) + C$$

In exercises 29 - 35, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

30) $$\displaystyle ∫\tan(2x)\,dx$$

Answer:
$$\displaystyle ∫\tan(2x)\,dx \quad = \quad −\frac{1}{2}\ln|\cos(2x)| + C \quad = \quad \frac{1}{2}\ln|\sec(2x)| + C$$
Solution:
\begin{align*} \displaystyle ∫\tan(2x)\,dx &= ∫\frac{\sin(2x)}{\cos(2x)}\,dx \\[5pt] &= -\frac{1}{2}\int\frac{1}{u}\,du & & \text{Letting } u = \cos(2x), \, \text{and}\, du = -2\sin(2x) dx \\[5pt] &= -\frac{1}{2}\ln|u| + C & & \text{Integrating in terms of}\, u \\[5pt] &=−\frac{1}{2}\ln|\cos(2x)| + C \quad = \quad \frac{1}{2}\ln|\sec(2x)| + C & & \text{Going back to}\,x \\[5pt] & & & \text{Gives us two equivalent forms of the antiderivative} \end{align*}

31) $$\displaystyle ∫\sec 5x\,dx$$

Answer:
$$\displaystyle ∫\sec 5x\,dx \quad = \quad \frac{1}{5}\ln|\tan 5x + \sec 5x | + C$$
Solution:
\begin{align*} \displaystyle ∫\sec 5x\,dx &= ∫\sec 5x\cdot\frac{\sec 5x + \tan 5x}{\sec 5x + \tan 5x}\,dx \\[5pt] &= \frac{1}{5} ∫\frac{5(\sec^2 5x + \sec 5x\tan 5x)}{\tan 5x + \sec 5x}\,dx \\[5pt] &= \frac{1}{5}\ln|\tan 5x + \sec 5x | + C \end{align*}

32) $$\displaystyle ∫\frac{x\sin(x^2)}{\cos(x^2)}\,dx$$

33) $$\displaystyle ∫\frac{\sin(3x)−\cos(3x)}{\sin(3x)+\cos(3x)}\,dx$$

Answer:
$$\displaystyle ∫\frac{\sin(3x)−\cos(3x)}{\sin(3x)+\cos(3x)}\,dx \quad = \quad −\frac{1}{3}\ln|\sin(3x)+\cos(3x)| + C$$

34) $$\displaystyle ∫x\csc(x^2)\,dx$$

Answer:
$$\displaystyle ∫x\csc(x^2)\,dx \quad = \quad −\frac{1}{2}\ln∣\csc(x^2)+\cot(x^2)∣+C$$

35) $$\displaystyle ∫\frac{e^x−e^{−x}}{e^x+e^{−x}}\,dx$$

Answer:
$$\displaystyle ∫\frac{e^x−e^{−x}}{e^x+e^{−x}}\,dx \quad = \quad \ln∣e^x+e^{−x}∣+C \quad = \quad \ln(e^x+e^{−x})+C$$

36) $$\displaystyle ∫\ln(\cos x)\tan x\,dx$$

37) $$\displaystyle ∫\ln(\csc x)\cot x\,dx$$

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

In exercises 38 - 39, $$f(x)≥0$$ for $$a≤x≤b$$. Find the area under the graph of $$f(x)$$ between the given values $$a$$ and $$b$$ by integrating.

38) $$f(x)=2^{−x};\quad a=1,b=2$$

39) $$f(x)=2^{−x};\quad a=3,b=4$$

Answer:
$$\dfrac{1}{\ln(65,536)}$$

40) Find the area under the graph of the function $$f(x)=xe^{−x^2}$$ between $$x=0$$ and $$x=5$$.

41) Compute the integral of $$f(x)=xe^{−x^2}$$ and find the smallest value of $$N$$ such that the area under the graph $$f(x)=xe^{−x^2}$$ between $$x=N$$ and $$x=N+10$$ is, at most, $$0.01$$.

Answer:
$$\displaystyle ∫^{N+1}_Nxe^{−x^2}\,dx=\frac{1}{2}(e^{−N^2}−e^{−(N+1)^2}).$$ The quantity is less than $$0.01$$ when $$N=2$$.

42) Find the limit, as $$N$$ tends to infinity, of the area under the graph of $$f(x)=xe^{−x^2}$$ between $$x=0$$ and $$x=5$$.

43) Show that $$\displaystyle ∫^b_a\frac{dt}{t}=∫^{1/a}_{1/b}\frac{dt}{t}$$ when $$0<a≤b$$.

Answer:
$$\displaystyle ∫^b_a\frac{dx}{x}=\ln(b)−\ln(a)=\ln(\frac{1}{a})−\ln(\frac{1}{b})=∫^{1/a}_{1/b}\frac{dx}{x}$$

44) Suppose that $$f(x)>0$$ for all $$x$$ and that $$f$$ and $$g$$ are differentiable. Use the identity $$f^g=e^{g\ln f}$$ and the chain rule to find the derivative of $$f^g$$.

45) Use the previous exercise to find the antiderivative of $$h(x)=x^x(1+\ln x)$$ and evaluate $$\displaystyle ∫^3_2x^x(1+\ln x)\,dx$$.

Answer:
23

46) Show that if $$c>0$$, then the integral of $$\frac{1}{x}$$ from $$ac$$ to $$bc$$ $$(\text{for}\,0<a<b)$$ is the same as the integral of $$\frac{1}{x}$$ from $$a$$ to $$b$$.

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition $$\displaystyle \ln(x)=∫^x_1\frac{dt}{t}$$, using properties of the definite integral and making no further assumptions.

47) Use the identity $$\displaystyle \ln(x)=∫^x_1\frac{dt}{t}$$ to derive the identity $$\ln(\dfrac{1}{x})=−\ln x$$.

Answer:
We may assume that $$x>1$$,so $$\dfrac{1}{x}<1.$$ Then, $$\displaystyle ∫^{1/x}_{1}\frac{dt}{t}$$. Now make the substitution $$u=\dfrac{1}{t}$$, so $$du=−\dfrac{dt}{t^2}$$ and $$\dfrac{du}{u}=−\dfrac{dt}{t}$$, and change endpoints: $$\displaystyle ∫^{1/x}_1\frac{dt}{t}=−∫^x_1\frac{du}{u}=−\ln x.$$

48) Use a change of variable in the integral $$\displaystyle ∫^{xy}_1\frac{1}{t}\,dt$$ to show that $$\ln xy=\ln x+\ln y$$ for $$x,y>0$$.

49) Use the identity $$\displaystyle \ln x=∫^x_1\frac{dt}{x}$$ to show that $$\ln(x)$$ is an increasing function of $$x$$ on $$[0,∞)$$, and use the previous exercises to show that the range of $$\ln(x)$$ is $$(−∞,∞)$$. Without any further assumptions, conclude that $$\ln(x)$$ has an inverse function defined on $$(−∞,∞).$$

50) Pretend, for the moment, that we do not know that $$e^x$$ is the inverse function of $$\ln(x)$$, but keep in mind that $$\ln(x)$$ has an inverse function defined on $$(−∞,∞)$$. Call it $$E$$. Use the identity $$\ln xy=\ln x+\ln y$$ to deduce that $$E(a+b)=E(a)E(b)$$ for any real numbers $$a$$, $$b$$.

51) Pretend, for the moment, that we do not know that $$e^x$$ is the inverse function of $$\ln x$$, but keep in mind that $$\ln x$$ has an inverse function defined on $$(−∞,∞)$$. Call it $$E$$. Show that $$E'(t)=E(t).$$

Answer:
$$x=E(\ln(x)).$$ Then, $$1=\dfrac{E'(\ln x)}{x}$$ or $$x=E'(\ln x)$$. Since any number $$t$$ can be written $$t=\ln x$$ for some $$x$$, and for such $$t$$ we have $$x=E(t)$$, it follows that for any $$t,\,E'(t)=E(t).$$

52) The sine integral, defined as $$\displaystyle S(x)=∫^x_0\frac{\sin t}{t}\,dt$$ is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large $$x$$. Show that for $$k≥1,\quad |S(2πk)−S(2π(k+1))|≤\dfrac{1}{k(2k+1)π}.$$ (Hint: $$\sin(t+π)=−\sin t$$)

53) [T] The normal distribution in probability is given by $$p(x)=\dfrac{1}{σ\sqrt{2π}}e^{−(x−μ)^2/2σ^2}$$, where $$σ$$ is the standard deviation and $$μ$$ is the average. The standard normal distribution in probability, $$p_s$$, corresponds to $$μ=0$$ and $$σ=1$$. Compute the left endpoint estimates $$R_{10}$$ and $$R_{100}$$ of $$\displaystyle ∫^1_{−1}\frac{1}{\sqrt{2π}}e^{−x^{2/2}}\,dx.$$

Answer:
$$R_{10}=0.6811,\quad R_{100}=0.6827$$

54) [T] Compute the right endpoint estimates $$R_{50}$$ and $$R_{100}$$ of $$\displaystyle ∫^5_{−3}\frac{1}{2\sqrt{2π}}e^{−(x−1)^2/8}$$.

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.

• Paul Seeburger (Monroe Community College) added problems #24 - 28 in 5.5 and #28 - 29, and 31 in 5.6.

5.5E and 5.6E u-Substitution Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

• Was this article helpful?