Checkpoint 1.19.
Hint.
See Example 1.17.
Appendix E Hints and Answers to Selected Exercises
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
Checkpoint 2.8.
Checkpoint 2.9.
Checkpoint 2.10.
II Calculus II
3 Techniques of Integration
3.1 Integration by Parts
Checkpoint 3.2.
Checkpoint 3.3.
3.2 Trigonometric Integrals
3.2.1 Products of Trigonometric Functions
Checkpoint 3.10.
Checkpoint 3.11.
3.2.2 Powers of Trigonometric Functions
Checkpoint 3.14.
Solution.
Let \(u = \sin(x)\text{.}\)
\begin{align*}
\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{align*}
Checkpoint 3.15.
Solution.
\begin{align*}
\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{align*}
Therefore,
\begin{align*}
\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{align*}
Checkpoint 3.16.
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{align*}
\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{align*}
\begin{align*}
\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{align*}
Putting this together,
\begin{align*}
\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{align*}
Solution 2.
We can also use Euler’s formula directly.
\begin{align*}
\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{align*}
\begin{align*}
\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{align*}
Checkpoint 3.21.
3.3 Trigonometric Substitutions
Checkpoint 3.32.
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{align*}
\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{align*}
Extending to the full domain by symmetry (as \(\frac{F(u)-F(-u)}{2}\) provides an odd antiderivative):
\begin{align*}
F(u)-F(-u) & =
\frac{1}{2}\left(\ln
\left|u+\sqrt{u^2-9}\right| -
\ln
\left|-u + \sqrt{(-u)^2-9}\right|\right) \\
&= \frac{1}{2}\ln
\left|
\frac{u+\sqrt{u^2-9}}{u-\sqrt{u^2-9}}\right| = \ln|u+\sqrt{u^2-9}|-\ln(3).
\end{align*}
Thus, the general solution is:
\begin{equation*}
\frac{1}{2}\ln|u+\sqrt{u^2-9}|+C = \frac{1}{2}\ln|2(x+1) +
\sqrt{4x^2+8x-5}| +C
\end{equation*}
4 Improper Integrals
4.1 Definitions and Examples
Checkpoint 4.2.
Checkpoint 4.5.
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 4.4.
Checkpoint 4.7.
Checkpoint 4.8.
Hint.
Checkpoint 4.9.
Hint.
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
Checkpoint 4.14.
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{align*}
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{align*}
which tends to \(\pi/2 - \pi/4 = \pi/4\) as \(t\) tends to \(\infty\text{.}\) This establishes the inequality.
Checkpoint 4.16.
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 4.12, 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.
