삼각치환

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

$1/8$

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).

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

Substitute $x=4\tan u$, $\mathrm{d}x=4 \sec^2u\,\mathrm{d}u$.

$$\begin{aligned} \int\frac{1}{x^2\sqrt{x^2+16}}\,\mathrm{d}x &=\int \frac{1}{16\tan^2 u \sqrt{16\tan^2 u+16}}\,4\sec^2 u\,\mathrm{d}u \\[0.1in] &=\int \frac{\sec^2u}{16\tan^2u\sec u}\,\mathrm{d}u =\frac{1}{16}\int \frac{\sec u}{\tan^2u}\,\mathrm{d}u \\[0.1in] &= \frac{1}{16}\int\frac{\cos u}{\sin^2 u}\,\mathrm{d}u \end{aligned}$$

To finish off the integral, we'll substitute $v=\sin u$, $\mathrm{d}v=\cos u\,\mathrm{d}u$.

$$\begin{aligned} \int\frac{1}{x^2\sqrt{x^2+16}}\,\mathrm{d}x &=\frac{1}{16} \int\frac{\cos u}{\sin^2 u}\,\mathrm{d}u = \frac{1}{16}\int\frac{\mathrm{d}v}{v^2} =-\frac{1}{16v} +C \\ &=-\frac{1}{16\sin u} +C =-\frac{1}{16}\dfrac{\sqrt{x^2+16}}{x}+C \smash{  } \end{aligned}$$

To find $\sin u$, we draw a right triangle with angle $u$ and $\tan u = \frac{x}{4}$. We label the opposite side $x$ and the adjacent side $4$, and then from Pythagoras we find that the hypotenuse has length $\sqrt{x^2+16}$. So, $\sin u = \dfrac{\sqrt{x^2+16}}{x}$.

As a check, we observe that the derivative of the answer

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left(-\frac{1}{16}\frac{\sqrt{x^2+16}}{x}+C\right) &=\frac{1}{16}\frac{\sqrt{x^2+16}}{x^2} -\frac{1}{16}\frac{x}{x\sqrt{x^2+16}} =\frac{1}{16}\frac{(x^2+16)-x^2}{x^2\sqrt{x^2+16}} \\ &=\frac{1}{x^2\sqrt{x^2+16}} \end{aligned}$$

is exactly the integrand.

Substitute $x=3\sec u$ with $0\le u<\frac{\pi}{2}$. Then $\mathrm{d}x= 3\sec u\tan u\,\mathrm{d}u$ and $\sqrt{x^2-9}=\sqrt{9\sec^2 u-9} =\sqrt{ 9\tan^2 u}=3\tan u$, so that

$$\begin{aligned} \int \frac{\mathrm{d}x}{x^2\sqrt{x^2-9}} &=\int \frac{3\sec u\tan u\ \mathrm{d}u} {9\sec^2u\sqrt{9\tan^2u}} \\ &=\frac{1}{9}\int\frac{\mathrm{d}u}{\sec u} \\ &=\frac{1}{9}\int \cos u\ \mathrm{d}u =\frac{1}{9}\sin u +C.\qquad \smash{  } \\ \text{ To evaluate $\sin u$, we make a right triangle with angle $u$. Since $\sec u = \dfrac{x}{3} = \dfrac{\text{hyp}}{\text{adj}}$, we label the hypotenuse $x$ and the adjacent side $3$. Using the Pythagorean Theorem, the opposite side has length $\sqrt{x^2-9}$. So, $\sin u = \dfrac{\sqrt{x^2-9}}x$ and} \\ \int \frac{\mathrm{d}x}{x^2\sqrt{x^2-9}} &= \frac{\sqrt{x^2-9}}{9x} +C. \end{aligned}$$

As a check, we observe that the derivative of the answer

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\sqrt{x^2-9}}{9x}+C\right) &=-\frac{\sqrt{x^2-9}}{9x^2} + \frac{x}{9x\sqrt{x^2-9}} =\frac{1}{9}\ \frac{-(x^2-9)+x^2}{x^2\sqrt{x^2-9}} \\ &=\frac{1}{x^2\sqrt{x^2-9}} \end{aligned}$$

is exactly the integrand. (We remark that this is the case even for $x\le -3$.)