삼각치환

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.