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,

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

If a quadratic function can be factored as $(ax+b)(cx+d)$ for some constants $a,b,c,d$, then it has roots $-\frac{b}{a}$ and $-\frac{d}{c}$. So, if a quadratic function has no roots, it is irreducible: this is the case for the function in graph (d).

If a quadratic function has two different roots, then $(ax+b) \neq \alpha(cx+d)$ for any constant $\alpha$. That is, the quadratic function is the product of distinct linear factors. This is the case for the functions graphed in (b) and (c), since these each have two distinct places where they cross the $x$-axis.

Finally, if a quadratic function has precisely one root, then $\frac{b}{a}=\frac{d}{c}$, so:

$$\begin{aligned} (ax+b)(cx+d)&=a(x+\tfrac{b}{a})(cx+d) = a(x+\tfrac{d}{c})(cx+d) = \tfrac{a}{c}(cx+d)(cx+d) \end{aligned}$$

That is, the quadratic function is the product of a repeated linear factor, and a constant $\frac{a}{c}$ (which might simply be $\frac{a}{c}=1$).

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: $$\approx 8.147\%$$

Midpoint rule:

Trapezoidal rule:

(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 $L=|-2|=2$.