Checkpoint 1.19.
Hint.
See Example 1.17.
Appendix F Hints and Selected Answers
I Calculus I
1 Differentiation
1.3 Derivatives of Elementary Functions
1.3.2 Derivatives of Exponential and Logarithmic Functions
2 Integration
2.2 Integration by Substitution
2.2.1 Overview
Checkpoint 2.8.
Hint.
Try the substitution \(u= 3x \text{.}\)
Solution.
Let \(u=3x\text{.}\) Then \(du =3 dx\) and \(\sin(3x) dx =
\frac{1}{3}\sin(u)du\text{.}\) So,
\begin{equation*}
\int \sin(3x)dx = \frac{1}{3}\int \sin(u) du = -\frac{1}{3}\cos(u)+C = -
\frac{1}{3}\cos(3x)+C
\end{equation*}
Checkpoint 2.9.
Hint.
Try the substitution \(u=e^x\text{.}\)
Solution.
Let \(u=e^x\text{.}\) Then \(du =e^xdx\) and \(1+e^{2x} = 1+u^2\text{.}\) So find the integral as
\begin{equation*}
\int \frac{du}{1+u^2} = \arctan(u) +C = \arctan(e^x) + C.
\end{equation*}
Checkpoint 2.10.
Hint.
Try the substitution \(u=\ln x\text{.}\)
II Calculus II
3 Techniques of Integration
3.1 Integration by Parts
3.1.1 Overview
Checkpoint 3.3.
Solution.
Set \(u=x\text{,}\) so \(du=dx\text{.}\) Let \(dv = \cos(x)\,dx\text{,}\) which implies \(v = \sin(x)\text{.}\) Applying IBP (3.1) yields:
\begin{equation*}
\begin{aligned}
\int x\cos(x)\,dx &= x\sin(x) - \int \sin(x)\,dx \\
&= x\sin(x) - (-\cos(x)) + C \\
&= x\sin(x) + \cos(x) + C.
\end{aligned}
\end{equation*}
Checkpoint 3.4.
Hint.
Set \(u=x^2\) and \(dv = \sin(x)\,dx\text{.}\) Then apply IBP twice.
Checkpoint 3.10.
Hint.
Use IBP with \(u=\arctan(1/x)\) and \(dv=9\,dx\text{.}\) Then simplify \(du\) and compute the resulting logarithmic integral.
Solution.
Let \(u=\arctan(1/x)\) and \(dv=9\,dx\text{.}\) Then \(v=9x\) and \(du= -\frac{1}{x^2+1}\,dx\) since \(\frac{d}{dx}\arctan(1/x)=\frac{-1/x^2}{1+(1/x)^2}=-\frac{1}{x^2+1}\text{.}\) By IBP,
\begin{equation*}
\begin{aligned}
\int_1^{\sqrt{3}} 9\arctan(1/x)\,dx &=
\left. 9x\arctan(1/x) \right|_1^{\sqrt{3}} - \int_1^{\sqrt{3}}
9x\left(-\frac{1}{x^2+1}\right)\,dx \\
&=
\left. 9x\arctan(1/x) \right|_1^{\sqrt{3}} + 9\int_1^{\sqrt{3}}
\frac{x}{x^2+1}\,dx \\
&= \left. 9x\arctan(1/x)
\right|_1^{\sqrt{3}} + \frac{9}{2}\left. \ln(x^2+1)
\right|_1^{\sqrt{3}}.
\end{aligned}
\end{equation*}
Evaluate the boundary terms:
\begin{equation*}
\arctan(1/\sqrt{3})=\pi/6,\quad
\arctan(1)=\pi/4,
\end{equation*}
so
\begin{equation*}
\begin{aligned}
\left. 9x\arctan(1/x) \right|_1^{\sqrt{3}} &=
9\left(\sqrt{3}\cdot \frac{\pi}{6} - \frac{\pi}{4}\right) \\
&= \frac{3\pi}{4}(2\sqrt{3}-3),
\end{aligned}
\end{equation*}
and
\begin{equation*}
\frac{9}{2}\left. \ln(x^2+1) \right|_1^{\sqrt{3}} =
\frac{9}{2}\ln\!\left(\frac{4}{2}\right)=\frac{9}{2}\ln 2.
\end{equation*}
Therefore,
\begin{equation*}
\int_1^{\sqrt{3}} 9\arctan(1/x)\,dx = \frac{3\pi}{4}(2\sqrt{3}-3)
+ \frac{9}{2}\ln 2.
\end{equation*}
3.2 Trigonometric Integrals
3.2.1 Products of Trigonometric Functions
Checkpoint 3.12.
Solution.
Apply the identity from (B.10).
\begin{equation*}
\begin{aligned}
\int \cos(3x)\cos(4x)dx &=
\frac{1}{2}\left(\int \cos(3x-4x)dx + \int \cos(3x+4x)dx \right) \\
&= \frac{1}{2}\sin(x) + \frac{1}{14}\sin(7x)+C.
\end{aligned}
\end{equation*}
Checkpoint 3.13.
Hint.
Recall the identity in (B.9).
Solution.
Set \(a=m+n\) and \(b=m-n\text{.}\)
If \(m\neq n\text{,}\) then
\begin{equation*}
\int \sin(mx)\cos(nx)\,dx
=-\dfrac{\cos((m+n)x)}{2(m+n)}-\dfrac{\cos((m-n)x)}{2(m-n)}+C.
\end{equation*}
If \(m=n\text{,}\) then
\begin{equation*}
\int \sin(mx)\cos(nx)\,dx = -\dfrac{1}{4m}\cos(2mx)+C.
\end{equation*}
3.2.2 Powers of Trigonometric Functions
Checkpoint 3.16.
Solution.
Let \(u = \sin(x)\text{.}\)
\begin{equation*}
\begin{aligned}
\int\sin^4(x)\cos^7(x)dx &=
\int u^4(\cos^2(x))^3\cos(x)dx \\
&= \int u^4 (1-u^2)^3\ du \\
&= \frac{1}{5}u^5 - \frac{3}{7}u^7 + \frac{3}{9}u^9 -
\frac{1}{11}u^{11} + C \\
& = \frac{1}{5}\sin^5 x - \frac{3}{7}\sin^7 x +
\frac{1}{3}\sin^9 x - \frac{1}{11}\sin^{11} x + C.
\end{aligned}
\end{equation*}
Checkpoint 3.17.
Solution.
\begin{equation*}
\begin{aligned}
\sin^4(x) &= (\sin^2(x))^2 =
\left(\frac{1-\cos(2x)}{2}\right)^2 \\
&= \frac{1}{4}\left(1 - 2\cos(2x) + \cos^2(2x)\right) \\
&= \frac{1}{4}\left(1 - 2\cos(2x) + \frac{1+\cos(4x)}{2}\right) \\
&= \frac{3}{8} - \frac{\cos(2x)}{2} + \frac{\cos(4x)}{8}
\end{aligned}
\end{equation*}
Therefore,
\begin{equation*}
\begin{aligned}
\int \sin^4(x)dx &= \int\frac{3}{8}dx -
\int\frac{\cos(2x)}{2}dx + \int\frac{\cos(4x)}{8}dx \\
&= \frac{3x}{8} - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C.
\end{aligned}
\end{equation*}
Checkpoint 3.18.
Solution 1.
Let \(w = 3x\text{.}\)
\begin{equation*}
\int \sin^2(3x)\cos^2(3x) dx = \frac{1}{3}\int\sin^2(w)\cos^2(w) dw.
\end{equation*}
\begin{equation*}
\begin{aligned}
\int \sin^2(w)\cos^2(w)dw &=
\int\frac{1-\cos(2w)}{2}\frac{1+\cos(2w)}{2} dw \\
& = \frac{1}{4} \int 1-\cos^2(2w)
\end{aligned}
\end{equation*}
Apply (B.14) to \(\cos^2(2w)\text{:}\)
\begin{equation*}
\begin{aligned}
\frac{1}{4} \int 1-\cos^2(2w) dw &=
\frac{1}{4} \int 1- \frac{1+\cos(4w)}{2} dw \\
&= \frac{1}{8}\int 1 - \cos(4w) dw \\
& = \frac{1}{8}\left(w - \frac{1}{4}\sin(4w)\right) + C \\
& = \frac{3x}{8} - \frac{\sin(12x)}{32} + C
\end{aligned}
\end{equation*}
Putting this together,
\begin{equation*}
\begin{aligned}
\int \sin^2(3x)\cos^2(3x) dx &=
\frac{1}{3}\int\sin^2(w)\cos^2(w) dw \\
& = \frac{1}{3}\left(\frac{3x}{8} - \frac{\sin(12x)}{32}\right)+C \\
& = \frac{x}{8} - \frac{\sin(12x)}{96} + C
\end{aligned}
\end{equation*}
Solution 2.
We can also use Euler’s formula directly.
\begin{equation*}
\begin{aligned}
\sin^2(3x)\cos^2(3x) &=
\left(\frac{e_{3x}-e_{-3x}}{2i}\right)^2
\left(\frac{e_{3x}+e_{-3x}}{2}\right)^2 \\
&=-\frac{1}{16}[(e_{3x}-e_{-3x})(e_{3x}+e_{-3x})]^2 \\
& = -\frac{1}{16}(e_{6x} -e_{-6x})^2 \\
& = \frac{1}{16}(2-(e_{12x}+e_{-12x})) \\
& = \frac{1}{8}\left(1 - \frac{e_{12x}+e_{-12x}}{2}\right) =
\frac{1}{8}(1-\cos(12x)).
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
\int \sin^2(3x)\cos^2(3x) dx &= \frac{1}{8} \int 1-\cos(12x) dx \\
& = \frac{x}{8} - \frac{\sin(12x)}{96} + C.
\end{aligned}
\end{equation*}
Checkpoint 3.25.
Hint.
Do it twice: (1) use \(u = \cos(x)\) and (2) use \(u=\tan(x)\text{.}\)
3.3 Trigonometric Substitutions
3.3.1 Overview
Checkpoint 3.33.
Solution.
Completing the square gives \(4x^2 + 8x -5= 4(x+1)^2 - 9\text{.}\) Let \(u=2(x+1)\text{.}\) The integral becomes:
\begin{equation*}
\frac{1}{2} \int \frac{1}{\sqrt{u^2-9}} du.
\end{equation*}
The integrand \(f(u) = 1/\sqrt{u^2-9}\) is even. If \(F(u)\) is an antiderivative for \(u \gt 3\text{,}\) then \(-F(-u)\) is one for \(u \lt -3\text{.}\) For \(u \gt 3\text{,}\) use \(u=3\sec(\theta)\text{:}\)
\begin{equation*}
\begin{aligned}
\frac{1}{2} \int \frac{3\sec(\theta)\tan(\theta)}{3\tan(\theta)} d\theta
& = \frac{1}{2}\int\sec(\theta) d\theta \\
& = \frac{1}{2}\ln|\sec(\theta)
+ \tan(\theta)|+C \\
& = \frac{1}{2}\ln
\left| \frac{u}{3} + \frac{\sqrt{u^2-9}}{3}\right| + C \\
& = \frac{1}{2}\ln
\left| u + \sqrt{u^2-9} \right| + C.
\end{aligned}
\end{equation*}
On the other component of the domain, the corresponding antiderivative has the opposite sign. Thus one antiderivative on each component is
\begin{equation*}
\begin{aligned}
\frac{1}{2}\operatorname{sgn}(u)\ln\left(|u|+
\sqrt{u^2-9}\right)+C \\
&= \,\frac{1}{2}\operatorname{sgn}(x+1)
\\
&\quad\ln\left(2|x+1|+\sqrt{4x^2+8x-5}\right)+C.
\end{aligned}
\end{equation*}
4 Improper Integrals
4.1 Definitions and Examples
4.1.1 Overview
Checkpoint 4.2.
Solution.
Graphs of \(1/\sqrt{x}, 1/x, 1/x^2\text{.}\)
\begin{equation*}
\int_1^t \frac{1}{x}dx = \ln|t| - \ln|1| = \ln t.
\end{equation*}
\begin{equation*}
\int_1^t \frac{1}{x^2}dx = \left. -\frac{1}{x} \right|_1^{t} =
\left.\frac{1}{x} \right|_t^1 = 1 - \frac{1}{t}.
\end{equation*}
and
\begin{equation*}
\int_1^t \frac{1}{\sqrt{x}}dx = \left. 2\sqrt{x} \right|_1^{t} =
2\sqrt{t}-2.
\end{equation*}
Therefore, as \(t \to +\infty\text{,}\) we have
\begin{equation*}
\int_1^{\infty} \frac{1}{x} dx = +\infty, \int_1^{\infty} \frac{1}{x^2} dx=
1, \int_1^{\infty}\\frac{1}{\sqrt{x}} dx = +\infty.
\end{equation*}
Checkpoint 4.5.
Hint.
Make the substitution \(u = 1/x\)
Solution.
For \(t \gt 0\text{,}\)
\begin{equation*}
\int_t^a f(x) dx \stackrel{u=1/x}{=} \int_{1/t}^{1/a}
f(1/u)\frac{-du}{u^2}
= \int_{1/a}^{1/t}\\frac{f(1/u)}{u^2}du.
\end{equation*}
As \(t \to 0^+, 1/t \to \infty\text{,}\) the first assertion follows. Apply that to \(f(x) = 1/x^p\text{,}\) we have
\begin{equation*}
\int_0^a \frac{1}{x^p} dx = \int_{1/a}^{\infty} \frac{1}{u^{2-p}}
du.
\end{equation*}
As a result, the second assertion follows from Proposition 4.3.
Checkpoint 4.8.
Hint.
Use integration by parts with \(u=\ln(x)\) and \(dv = x^p dx\text{.}\)
Solution.
For \(a\in(0,1)\text{,}\) let
\begin{equation*}
I(a)=\int_a^1 x^p\ln(x)\,dx.
\end{equation*}
Using integration by parts with \(u=\ln(x)\) and \(dv=x^p\,dx\text{,}\) for \(p\neq -1\) we get
\begin{equation*}
\begin{aligned}
I(a)
&=
\left.\frac{x^{p+1}}{p+1}\ln(x)\right|_a^1
-\frac{1}{p+1}\int_a^1 x^p\,dx \\
&=
-\frac{a^{p+1}\ln(a)}{p+1}
-\frac{1-a^{p+1}}{(p+1)^2}.
\end{aligned}
\end{equation*}
If \(p\gt -1\text{,}\) then \(a^{p+1}\to 0\) and \(a^{p+1}\ln(a)\to 0\) as \(a\to 0^+\text{.}\) Hence
\begin{equation*}
\int_0^1 x^p\ln(x)\,dx
=\lim_{a\to 0^+} I(a)
=-\frac{1}{(p+1)^2}.
\end{equation*}
If \(p\lt -1\text{,}\) then \(a^{p+1}\to\infty\text{,}\) so the expression above tends to \(-\infty\text{;}\) thus the integral diverges. For \(p=-1\text{,}\) \(\int_a^1 \frac{\ln(x)}{x}\,dx=-\frac{1}{2}(\ln(a))^2\to-\infty\text{,}\) so the original integral diverges (to \(-\infty\)).
Therefore, the improper integral converges exactly when \(p\gt -1\text{,}\) and in that case
\begin{equation*}
\int_0^1 x^p\ln(x)\,dx=-\frac{1}{(p+1)^2}.
\end{equation*}
Checkpoint 4.9.
Hint.
Use the substitution \(u=\ln(x)\) and then use Proposition 4.3
Checkpoint 4.10.
Hint.
First, sketch the region. Then recognize the area is given by the integral
\begin{equation*}
\int_{-2}^0 \frac{1}{\sqrt{x+2}} dx.
\end{equation*}
To find the integral, make the substitution \(u=x+2\) then use the result in Checkpoint 4.5.
4.2 Comparison Test for Integrals
4.2.1 Overview
Checkpoint 4.15.
Solution.
Since \(0 \le \cos^2(x) \le 1\text{,}\) \(0 \le
\frac{\cos^2(x)}{1+x^2} \le \frac{1}{1+x^2}\text{,}\) it follows from the comparison test for integrals that
\begin{equation*}
\begin{aligned}
0 & \le \int_1^{t} \frac{\cos^2(x)}{1+x^2}dx
\le \int_1^{t} \frac{1}{1+x^2}dx \\
& =
\arctan(t)-\frac{\pi}{4}
\end{aligned}
\end{equation*}
which tends to \(\pi/2 - \pi/4 = \pi/4\) as \(t\) tends to \(\infty\text{.}\) This establishes the inequality.
Checkpoint 4.17.
Hint.
The integrand is
\begin{equation*}
\frac{(3-C)x^2 +x - C}{(x^2+1)(3x+1)}
\end{equation*}
By degree consideration and Proposition 4.13, we know that \(C\) must be \(3\text{.}\) So original integrand must be
\begin{equation*}
\frac{x}{x^2+1} - \frac{3}{3x+1}
\end{equation*}
Now find an anti-derivative of each term above then find the value of the improper integral by computing suitable limits.
5 Sequence and Series
5.1 Sequences
5.1.2 Algebraic Operations on Sequences
Checkpoint 5.7.
Solution.
The \(n\)-th of their sum is \(n^2 + 1/n\) and that of their product is \((n^2)(1/n) = n\text{.}\)
5.5 Power Series
5.5.1 Examples and Exercises
Checkpoint 5.68.
Hint.
Note that
\begin{equation*}
f(x) = \frac{3}{5}\frac{1}{1+x/5}.
\end{equation*}
Then substitute \(u=-x/5\) into the expansion of \(1/(1-u)\text{.}\)
Checkpoint 5.69.
Hint.
Task. Find the coefficients \(a_n\) in
\begin{equation*}
\cos(x) = \sum_{n=0}^{\infty} a_n(x-\pi/2)^n
\end{equation*}
Note that
\begin{equation*}
\cos(x + \pi/2) = \sum_{n=0}^{\infty} a_n(x+\pi/2 - \pi/2)^n =
\sum_{n=0}^{\infty} a_n x^n
\end{equation*}
\(\cos(x+\pi/2) = \cos(x)\cos(\pi/2)-\sin(x)\sin(\pi/2)
= -\sin(x)\text{.}\) Then use the power series expansion of \(sin(x)\) at \(x_0=0\) to find the \(a_n\)’s.
