적분 기법 전체 문제

If $u=\cos x$, then $\mathrm{d}u=-\sin x\,\mathrm{d}x$. If $n \neq -1$, then

$$\begin{aligned} \int_0^{\pi/4} \sin x \cos^n x ~\mathrm{d}x&= - \int_1^{1/\sqrt2} u^n \mathrm{d}u = \left[-\frac{1}{n+1}u^{n+1}\right]_1^{1/\sqrt2}=\frac{1}{n+1}\left(1-\frac{1}{\sqrt{2}^{n+1}}\right) \\ \text{If $n=-1$, then} \\ \int_0^{\pi/4} \sin x \cos^n x ~\mathrm{d}x&= - \int_1^{1/\sqrt2} u^n \mathrm{d}u = - \int_1^{1/\sqrt2} \frac{1}{u} \mathrm{d}u =\bigg[ -\log|u|\bigg]_1^{1/\sqrt2} \\ & =-\log\left(\frac{1}{\sqrt2}\right) = \frac{1}{2}\log 2 \end{aligned}$$

So, (e) $n$ can be any real number.

We use the substitution $u=\sec x$, $\mathrm{d}u=\sec x \tan x ~\mathrm{d}x$.

$$\begin{aligned} \int \sec^n x \tan x \mathrm{d}x&=\int \sec^{n-1}x \cdot \sec x \tan x~\mathrm{d}x = \int u^{n-1}\mathrm{d}u \\ \text{Since $n$ is positive, $n-1 \neq -1$, so we antidifferentiate using the power rule.} \\ &=\frac{u^n}{n}+C = \frac{1}{n}\sec^n x +C \end{aligned}$$

We divide both sides by $\cos^2 x$, and simplify.

$$\begin{aligned} \sin^2x+\cos^2 x &=1 \\\frac{\sin^2x+\cos^2 x }{\cos^2 x}&=\frac{1}{\cos^2 x} \\\frac{\sin^2x}{\cos^2 x}+1&=\sec^2 x \\\tan^2 x+1&=\sec^2 x \end{aligned}$$

In the text, there is a template for choosing an appropriate substitution, but for this problem we will explain the logic of the choices.

The trig identities that we can use are:

$$\begin{aligned} &1-\sin^2\theta=\cos^2\theta & &\tan^2 \theta +1 = \sec^2 \theta & &\sec^2 \theta - 1 =\tan^2 \theta \\ \text{They have the following forms:} \\ &\text{constant } - \text{ function} &&\text{function } + \text{ constant} & &\text{function } - \text{ constant} \end{aligned}$$

In order to cancel out the square root, we should choose a substitution that will match the argument under the square root with the trig identity of the corresponding form.

(a) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function $9x^2-16$, which has the form (function) $-$ (constant). This form matches the trig identity $\sec^2 \theta - 1 = \tan^2 \theta$. We can set $x$ to be whatever we need it to be, but we don't have the same control over the constant, 16. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 16, we get

$$\begin{aligned} ~16\sec^2\theta - 16 &= 16\tan^2\theta \\ \text{What we want is a substitution that gives us} \\ 9x^2-16&=16\sec^2\theta - 16\\ \text{So,}\qquad 9x^2&=16\sec^2\theta\\ x &= \frac{4}{3}\sec\theta \\ \text{Using this substitution,} \\ \sqrt{9x^2-16}&=\sqrt{16\sec^2\theta-16}\\ &=\sqrt{16\tan^2\theta}\\ &=4|\tan\theta| \end{aligned}$$

So, we eliminated the square root.

(b) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function $1-4x^2$, which has the form (constant) $-$ (function). This form matches the trig identity $1-\sin^2\theta = \cos^2\theta$. What we want is a substitution that gives us

$$\begin{aligned} 1-4x^2&=1-\sin^2\theta\\ \text{So,}\qquad 4x^2&=\sin^2\theta\\ x &= \frac{1}{2}\sin\theta \\ \text{Using this substitution,} \\ \sqrt{1-4x^2}&=\sqrt{1-\sin^2\theta}\\ &=\sqrt{\cos^2\theta}\\ &=|\cos\theta| \end{aligned}$$

So, we eliminated the square root. We remark that the absolute value signs are not needed in $|\cos\theta|$, because, for $-\frac{1}{2}\le x\le\frac{1}{2}$, we have $\theta=\arcsin(2x)$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and $\cos(\theta)\ge 0$ for those $\theta$'s.

(c) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the fractional power. (That is, we want to eliminate the square root.) The function under the power is $25+x^2$, which has the form (constant) $+$ (function). This form matches the trig identity $\tan^2 \theta + 1 = \sec^2 \theta$. We can set $x$ to be whatever we need it to be, but we don't have the same control over the constant, 25. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 25, we get

$$\begin{aligned} 25\tan^2 \theta + 25 &= 25\sec^2 \theta \\ \text{What we want is a substitution that gives us} \\ 25+x^2&=25\tan^2\theta+25\\ \text{So,}\qquad x^2&=25\tan^2\theta\\ x &= 5\tan\theta \\ \text{Using this substitution,} \\ (25+x^2)^{-5/2}&=(25+25\tan^2\theta)^{-5/2}\\ &=(25\sec^2\theta)^{-5/2}\\ &=(5|\sec\theta|)^{-5} \end{aligned}$$

So, we eliminated the square root. We remark that the absolute value signs are not needed in $|\sec\theta|$, because, for $-\infty< x<\infty$, we have $\theta=\arctan(x/5)$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and $\sec(\theta)\ge 0$ for those $\theta$'s.

(a) If $\sin\theta=\dfrac{1}{20}$ and $\theta$ is between 0 and $\pi/2$, then we can draw a right triangle with angle $\theta$ that has opposite side length 1, and hypotenuse length 20. By the Pythagorean Theorem, the adjacent side has length $\sqrt{20^2-1^2}=\sqrt{399}$. So, $\cos\theta = \dfrac{\mathrm{adj}}{\mathrm{hyp}}=\dfrac{\sqrt{399}}{20}$.

We can do a quick “reasonableness" check here: $\frac{1}{20}$ is pretty close to 0, so we might expect $\theta$ to be pretty close to 0, and so $\cos \theta$ should be pretty close to 1. Indeed it is: $\dfrac{\sqrt{399}}{20}\approx \dfrac{\sqrt{400}}{20}=\dfrac{20}{20}=1$.

Alternately, we can solve this problem using identities.

$$\begin{aligned} \sin^2 \theta + \cos^2 \theta &=1\\ \left(\frac{1}{20}\right)^2+ \cos^2 \theta &=1\\ \cos\theta &= \pm\sqrt{1-\frac{1}{400}}=\pm\frac{\sqrt{399}}{20} \\ \text{Since $0 \leq \theta \leq \frac{\pi}{2}$, $\cos\theta \geq 0$, so} \\ \cos\theta &= \frac{\sqrt{399}}{20} \end{aligned}$$ (b) If $\tan\theta=7$ and $\theta$ is between 0 and $\pi/2$, then we can draw a right triangle with angle $\theta$ that has opposite side length 7 and adjacent side length 1. By the Pythagorean Theorem, the hypotenuse has length $\sqrt{7^2+1^2} = \sqrt{50}=5\sqrt{2}$. So, $\csc\theta = \dfrac{\mathrm{hyp}}{\mathrm{opp}}=\dfrac{5\sqrt{2}}{7}$.

Again, we can do a quick reasonableness check. Since 7 is much larger than 1, the triangle we're thinking of doesn't look much like the triangle in our standardized picture above: it's really quite tall, with a small base. So, the opposite side and hypotenuse are pretty close in length. Indeed, $\dfrac{5\sqrt{2}}{7}\approx 7.071$, so this dimension seems reasonable. (c) If $\sec\theta=\dfrac{\sqrt{x-1}}{2}$ and $\theta$ is between 0 and $\pi/2$, then we can draw a right triangle with angle $\theta$ that has hypotenuse length $\sqrt{x-1}$ and adjacent side length 2. By the Pythagorean Theorem, the opposite side has length $\sqrt{\sqrt{x-1}^2 - 2^2} = \sqrt{x-1-4}=\sqrt{x-5}$. So, $\tan\theta = \dfrac{\mathrm{opp}}{\mathrm{adj}}=\dfrac{\sqrt{x-5}}{2}$.

We can also solve this using identities. Note that since $\sec\theta$ exists, $\theta \neq \frac{\pi}{2}$.

$$\begin{aligned} \tan^2\theta+1&=\sec^2\theta\\ \tan^2\theta+1&=\left(\frac{\sqrt{x-1}}{2}\right)^2=\frac{x-1}{4}\\ \tan\theta &= \pm\sqrt{\frac{x-1}{4}-1} = \pm\frac{\sqrt{x-5}}{2} \\ \text{Since $0 \leq \theta < \frac{\pi}{2}$, $\tan\theta \geq 0$, so} \\ \tan\theta &= \frac{\sqrt{x-5}}{2} \end{aligned}$$

(a) Let $\theta = \arccos \left(\frac{x}{2}\right)$. That is, $\cos(\theta) = \frac{x}{2}$, and $0 \leq \theta \leq \pi$. Then we can draw the corresponding right triangle with angle $\theta$ with adjacent side of signed length $x$ (we note that if $\theta > \frac{\pi}{2}$, then $x$ is negative) and hypotenuse of length $2$. By the Pythagorean Theorem, the opposite side of the triangle has length $\sqrt{4-x^2}$.

So,

$$\sin\left(\arccos \left(\frac{x}{2}\right)\right)=\sin \theta = \frac{\mathrm{opp}}{\mathrm{hyp}} = \frac{\sqrt{4-x^2}}{2}$$ (b) Let $\theta = \arctan \left(\frac{1}{\sqrt{3}}\right)$. That is, $\tan(\theta) = \frac{1}{\sqrt{3}}$, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

• Solution 1: Then $\theta = \dfrac{\pi}{6}$, so $\sin\theta = \dfrac{1}{2}$.
• Solution 2: Then we can draw the corresponding right triangle with angle $\theta$ with opposite side of length $1$ and adjacent side of length $\sqrt{3}$. By the Pythagorean Theorem, the hypotenuse of the triangle has length $\sqrt{\sqrt{3}^2+1^2}=2$.

So,

$$\sin\left(\arctan \left(\frac{1}{\sqrt{3}}\right)\right)=\sin \theta = \frac{\mathrm{opp}}{\mathrm{hyp}} = \frac{1}{2}$$ (c) Let $\theta = \arcsin \left(\sqrt{x}\right)$. That is, $\sin(\theta) = \sqrt{x}$, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. Then we can draw the corresponding right triangle with angle $\theta$ with opposite side of length $\sqrt{x}$ and hypotenuse of length $1$. By the Pythagorean Theorem, the adjacent side of the triangle has length $\sqrt{1-x}$.

So,

$$\sec\left(\arcsin \left(\sqrt{x}\right)\right)=\sec \theta = \frac{\mathrm{hyp}}{\mathrm{adj}} = \frac{1}{\sqrt{1-x}}$$

The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution $u=(ax+b)$. It's less clear how to find the antiderivative of (2).

The absolute error is the difference between the two values: $$|1.387-1.5| = 0.113$$ The relative error is the absolute error divided by the exact value: $$\frac{0.113}{1.387}\approx 0.08147$$ The percent error is 100 times the relative error: $$\approx 8.147\%$$

(a) Differentiating, we find $f''(x) = -x^2+7x-6$. Since $f''(x)$ is quadratic, we have a pretty good idea of what it looks like.

• It factors as $f(x) = -(x-6)(x-1)$, so its two roots are at $x=6$ and $x=1$.
• The “flat part" of the parabola is at $x=3.5$ (since this is exactly half way between $x=1$ and $x=6$; alternately, we can check that $f'''(3.5)=0$).
• Since the coefficient of $x^2$ is negative, $f(x)$ is increasing from $-\infty$ to $3.5$, then decreasing from $3.5$ to $\infty$.

Therefore, over the interval $[1,6]$, the largest positive value of $f''(x)$ occurs when $x=3.5$, and this is $f''(3.5) = -(3.5-6)(3.5-1)=6.25$.

So, we take $M=6.25$. (b) We differentiate further to find $f^{(4)}(x)=-2$. This is constant everywhere, so we take $L=|-2|=2$.

Under any reasonable assumptions (Anyone caught trying to come up with a scenario in which I am older than my mother will be sent to maximum security grad school.), my mother is older than I am.

If $b= \pm \infty$, then our integral is improper because one limit is not a real number.

Furthermore, our integral will be improper if its domain of integration contains either of its infinite discontinuities, $x=1$ and $x=-1$. Since one limit of integration is 0, the integral is improper if $b \geq 1$ or if $b \leq -1$.

Below, we've graphed $\frac{1}{x^2-1}$ to make it clearer why values of $b$ in $(-1,1)$ are the only values that don't result in an improper integral when the other limit of integration is $a=0$.

Since the integrand is continuous for all real $x$, the only kind of impropriety available to us is to set $b = \pm\infty$.