삼각치환

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 $x\le -3$.)

(a) We'll use the trig identity $\cos2\theta=2\cos^2\theta-1$. It implies that

$$\begin{aligned} \cos^2\theta=\frac{\cos2\theta+1}{2} \implies \cos^4\theta &=\frac{1}{4}\big[\cos^22\theta+2\cos2\theta+1\big] =\frac{1}{4}\Big[\frac{\cos4\theta+1}{2}+2\cos2\theta+1\Big]\\ &=\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8} \\ \text{So,} \\ \int_0^{\pi/4}\cos^4\theta\ \mathrm{d}\theta &=\int_0^{\pi/4}\Big(\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8}\Big) \ \mathrm{d}\theta \\ &=\left[\frac{\sin4\theta}{32}+\frac{\sin2\theta}{4}+\frac{3}{8}\theta\right]_0^{\pi/4}\\ &= \frac{1}{4}+\frac{3}{8}\cdot \frac{\pi}{4}\\ &=\frac{8+3\pi}{32} \end{aligned}$$

as required.

(b) We'll use the trig substitution $x=\tan\theta$, $\mathrm{d}x=\sec^2\theta\ \mathrm{d}\theta$. Note that when $\theta=\pm\frac{\pi}{4}$, we have $x=\pm 1$. Also note that dividing the trig identity $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ gives the trig identity $\tan^2\theta+1=\sec^2\theta$. So

$$\begin{aligned} \int_{-1}^1\frac{\mathrm{d}x}{{(x^2+1)}^3} &=2\int_0^1\frac{\mathrm{d}x}{{(x^2+1)}^3}&\text{(even integrand)}\\ &=2\int_0^{\pi/4}\frac{\sec^2\theta\ \mathrm{d}\theta}{{(\tan^2\theta+1)}^3}\\ &=2\int_0^{\pi/4}\frac{\sec^2\theta\ \mathrm{d}\theta}{{(\sec^2\theta)}^3}\\ &=2\int_0^{\pi/4}\cos^4\theta\ \mathrm{d}\theta\\ &=\frac{8+3\pi}{16} \end{aligned}$$

by part (a).

The integrand is an odd function, and the limits of integration are symmetric, so $\displaystyle\int_{-\pi/12}^{\pi/12} \dfrac{15x^3}{(x^2+1)\sqrt{9-x^2}^5}~\mathrm{d}x=0$.

Substitute $x=\frac{2}{5}\sec u$ with $0< u<\frac{\pi}{2}$, so that $\mathrm{d}x=\frac{2}{5} \sec u\,\tan u\,\mathrm{d}u$ and $\sqrt{25 x^2-4} = \sqrt{4(\sec^2u-1)} = \sqrt{4\tan^2u}=2\tan u$. Then

$$\begin{aligned} \int \frac{\sqrt{25x^2-4}}{x}\,\mathrm{d}x &=\int \frac{2\tan u}{\frac{2}{5}\sec u}\cdot\frac{2}{5}\sec u\tan u\,\mathrm{d}u \\ &=2\int \tan^2 u\,\mathrm{d}u =2\int \big(\sec^2 u -1\big)\,\mathrm{d}u \\ &= 2\tan u -2u + C \\ &=\sqrt{25x^2-4}-2\mathop{\mathrm{arcsec}}\tfrac{5x}{2} + C \smash{  } \end{aligned}$$

To find $\tan u$, we draw a right triangle with angle $u$. Since $\sec u =\dfrac{5x}{2}$, we label the hypotenuse $5x$ and the adjacent side 2. Then the Pythagorean Theorem gives us the opposite side as length $\sqrt{25x^2-4}$. Then $\tan u = \dfrac{\text{opp}}{\text{adj}} = \dfrac{\sqrt{25x^2-4}}{2}$.

Alternately, we can notice that in our work, we already showed $2\tan u = \sqrt{25x^2-4}$, so $\tan u = \frac{1}{2}\sqrt{25x^2-4} .$

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

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left(\sqrt{25x^2-4}-2\mathop{\mathrm{arcsec}}\frac{5x}{2} + C\right) &=\frac{25 x}{\sqrt{25x^2-4}} - 2 \frac{\frac{5}{2}}{\left|\frac{5x}{2}\right|\sqrt{\frac{25x^2}{4}-1}} \\ &= \frac{25 x}{\sqrt{25x^2-4}} - \frac{4}{x\sqrt{25x^2-4}}\qquad\text{since $x>0$} \\ &= \frac{25 x^2-4}{x\sqrt{25x^2-4}} \\ &= \frac{\sqrt{25 x^2-4}}{x} \end{aligned}$$

is exactly the integrand (provided $x>\frac{2}{5}$).

The integrand has a quadratic polynomial under a square root, which makes us think of trig substitutions. However, it's good practice to look for simpler methods before we jump into more complicated ones, and in this case we find something nicer than a trig substitution: the substitution $u=x^2-1$, $\mathrm{d}u=2x\,\mathrm{d}x$. Then $x\mathrm{d}x = \frac{1}{2}\mathrm{d}u$, and $x^2 ={u+1}$. When $x=\sqrt{10}$, $u=9$, and when $x=\sqrt{17}$, $u=16$.

$$\begin{aligned} \int_{\sqrt{10}}^{\sqrt{17}} \frac{x^3}{\sqrt{x^2-1}}\, \mathrm{d}x&= \int_{\sqrt{10}}^{\sqrt{17}} \frac{x^2}{\sqrt{x^2-1}}\, \cdot x\mathrm{d}x \\&=\frac{1}{2} \int_{9}^{16} \frac{u+1}{\sqrt{u}}\,\mathrm{d}u\\ &=\frac{1}{2} \int_{9}^{16}\left(u^{1/2}+u^{-1/2}\right)\,\mathrm{d}u \\&=\frac{1}{2}\left[\frac{2}{3}u^{3/2} + 2u^{1/2} \right]_{9}^{16} \\&=\frac{1}{2}\left[\frac{2}{3}\cdot 4^3 + 2\cdot 4 -\frac{2}{3}\cdot 3^3 -2\cdot 3 \right]\\ &=\frac{40}{3} \end{aligned}$$

This integrand looks very different from those above. But it is only slightly disguised. If we complete the square

$$\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{3-2x-x^2}} = \int \frac{\mathrm{d}x}{\sqrt{4-(x+1)^2}} \end{aligned}$$

and make the substitution $y=x+1$, $\mathrm{d}y=\mathrm{d}x$

$$\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{3-2x-x^2}} = \int \frac{\mathrm{d}x}{\sqrt{4-(x+1)^2}} = \int \frac{\mathrm{d}y}{\sqrt{4-y^2}} \end{aligned}$$

we get a typical trig substitution integral. So, we substitute $y=2\sin\theta$, $\mathrm{d}y=2\cos\theta\,\mathrm{d}\theta$ to get

$$\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{3-2x-x^2}} &= \int \frac{\mathrm{d}y}{\sqrt{4-y^2}} = \int\frac{2\cos\theta\,\mathrm{d}\theta}{\sqrt{4-4\sin^2\theta}} = \int\frac{2\cos\theta\,\mathrm{d}\theta}{\sqrt{4\cos^2\theta}}\\ &=\int\mathrm{d}\theta =\theta +C =\arcsin\frac{y}{2} + C \\ &=\arcsin\frac{x+1}{2} + C \end{aligned}$$

An experienced integrator would probably substitute $x+1 = 2\sin\theta$ directly, without going through $y$.

Completing the square, we see $4x^2-12x+8 = (2x-3)^2-1$.

$$\begin{aligned} \int \dfrac{1}{(2x-3)^3\sqrt{4x^2-12x+8}}~\mathrm{d}x&= \int \dfrac{1}{(2x-3)^3\sqrt{(2x-3)^2-1}}~\mathrm{d}x \\ \text{As $x>2$, we have $2x-3>1$. We use the substitution $2x-3 = \sec \theta$ with $0\le\theta<\frac{\pi}{2}$. So $2\,\mathrm{d}x=\sec\theta\tan\theta~\mathrm{d}\theta$ and $\sqrt{(2x-3)^2-1}=\sqrt{\sec^2\theta-1}=\sqrt{\tan^2\theta}=\tan\theta$.} \\ &=\frac{1}{2}\int\frac{1}{\sec^3\theta\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta~\mathrm{d}\theta\\ &=\frac{1}{2}\int\frac{1}{\sec^3\theta\tan\theta}\sec\theta\tan\theta~\mathrm{d}\theta \\&=\frac{1}{2}\int\frac{1}{\sec^2\theta}~\mathrm{d}\theta \\&=\frac{1}{2}\int{\cos^2\theta}~\mathrm{d}\theta \\&=\frac{1}{4}\int{\left(1+\cos(2\theta)\right)}~\mathrm{d}\theta \\&=\frac14\left(\theta + \frac{1}{2}\sin(2\theta)\right)+C \\&=\frac14\left(\theta + \sin\theta\cos\theta\right)+C \\ \smash{  }\,&=\frac{1}{4}\left(\arccos\left(\frac{1}{2x-3}\right) + \frac{\sqrt{4x^2-12x+8}}{(2x-3)^2}\right)+C \end{aligned}$$