삼각치환

$x+\frac{x^2}{2}+\frac{125}{9} \ln |x-5|+\frac{64}{9} \ln |x+4|+C$

$2x+C$

$\frac{1}{6} \left(-\ln \left|x^2+2 x+3\right|+2 \ln |x|-\sqrt{2} \tan ^{-1}\left(\frac{x+1}{\sqrt{2}}\right)\right)+C$

$-\frac{3}{2} \ln \left|x^2+4 x+10\right|+x+\frac{\tan \ ^{-1}\left(\frac{x+2}{\sqrt{6}}\right)}{\sqrt{6}}+C$

$\ln \left|3 x^2+5 x-1\right|+2 \ln |x+1|+C$

$2\ln |x-3| +2\ln |x^2+6x+10| -4\tan^{-1}(x+3) +C$

$\frac{9}{10} \ln \left|x^2+9\right|+\frac{1}{5} \ln |x+1|-\frac{4}{15} \tan ^{-1}\left(\frac{x}{3}\right) +C$

$\frac{1}{2} \left(3 \ln \left|x^2+2 x+17\right|-4 \ln |x-7|+\tan \ ^{-1}\left(\frac{x+1}{4}\right)\right) +C$

$3 \left(\ln \left|x^2-2 x+11\right|+\ln |x-9|\right)+3 \ \sqrt{\frac{2}{5}} \tan ^{-1}\left(\frac{x-1}{\sqrt{10}}\right) +C$

$\frac{1}{2} \ln \left|x^2+10 x+27\right|+5 \ln |x+2|-6 \sqrt{2} \tan \ ^{-1}\left(\frac{x+5}{\sqrt{2}}\right) +C$

$\ln (2000/243) \approx 2.108$

$5\ln (9/4)-\frac13\ln (17/2) \approx 3.3413$

$-\pi/4+\tan^{-1} 3 - \ln (11/9) \approx 0.263$

$1/8$

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 quadratic formula underneath the square root makes us think of a trig substitution, but in the interest of developing good habits, let's check for an easier way first. If we let $u=2x^2+4x$, then $\mathrm{d}u = (4x+4)~\mathrm{d}x$, so $\frac{1}{4}\,\mathrm{d}u=(x+1)\,\mathrm{d}x$. This substitution looks easier than a trig substitution (which would start with completing the square).

$$\begin{aligned} \int\frac{x+1}{\sqrt{2x^2+4x}} \, \mathrm{d}x&=\frac{1}{4}\int \frac{1}{\sqrt{u}}\,\mathrm{d}u = \frac{1}{2}\sqrt{u}+C = \frac{1}{2}\sqrt{2x^2+4x}+C \end{aligned}$$