삼각치환

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 $4x^2-12x+8 = (2x-3)^2-1$.

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

Since $2x-3=\sec\theta$, we know $\cos\theta = \frac{1}{2x-3}$ and $\theta = \arccos\left(\frac{1}{2x-3}\right)$. (Equivalently, $\theta = \mathop{\mathrm{arcsec}}(2x-3)$.) To find $\sin\theta$, we draw a right triangle with adjacent side of length 1, and hypotenuse of length $2x-3$. By the Pythagorean Theorem, the opposite side has length $\sqrt{4x^2-12x+8}$.

We use the substitution $x=\tan u$, $\mathrm{d}x=\sec^2 u~\mathrm{d}u$. Note $\tan 0=0$ and $\tan \frac{\pi}{4} =1$.

$$\begin{aligned} \displaystyle\int_0^1\dfrac{x^2}{\sqrt{x^2+1}^3}\mathrm{d}x&= \int_0^{\pi/4}\dfrac{\tan^2 u}{\sqrt{\tan^2 u +1}^3}\sec^2u~\mathrm{d}u\\ &= \int_0^{\pi/4}\dfrac{\tan^2 u}{\sqrt{\sec^2u}^3}\sec^2u~\mathrm{d}u\\ &= \int_0^{\pi/4}\frac{\tan^2 u}{\sec u}~\mathrm{d}u\\&= \int_0^{\pi/4}\frac{\sec^2 u-1}{\sec u}~\mathrm{d}u\\&= \int_0^{\pi/4}\big({\sec u}-\cos u\big)~\mathrm{d}u\\ &=\Big[\log\left|\sec u + \tan u \right| - \sin u\Big]_{0}^{\pi/4} \\ &=\left(\log\left|\sqrt{2} + 1 \right| -\frac{1}{\sqrt{2}}\right) - \left(\log|1+0|-0\right)\\ &=\log(1+\sqrt{2})-\frac{1}{\sqrt{2}} \end{aligned}$$

There's no square root, but we can still make use of the substitution $x=\tan\theta$, $\mathrm{d}x = \sec^2\theta~\mathrm{d}\theta$.

$$\begin{aligned} \int \frac{1}{(x^2+1)^2}~\mathrm{d}x&=\int\frac{1}{(\tan^2\theta+1)^2}\sec^2\theta~\mathrm{d}\theta\\ &=\int\frac{1}{\sec^4\theta}\sec^2\theta~\mathrm{d}\theta = \int\cos^2\theta~\mathrm{d}\theta\\ &=\frac{1}{2}\int \big(1 + \cos(2\theta)\big)~\mathrm{d}\theta\\ &=\frac{1}{2}\left(\theta + \frac{1}{2}\sin(2\theta)\right)+C\\ &=\frac{1}{2}\left(\theta + \sin\theta\cos\theta\right)+C\\ \smash{  }\,&=\frac{1}{2}\left(\arctan x + \frac{x}{x^2+1}\right)+C \end{aligned}$$

Since $x = \tan\theta$, we can draw a right triangle with angle $\theta$, opposite side $x$, and adjacent side $1$. Then by the Pythagorean Theorem, its hypotenuse has length $\sqrt{x^2+1}$, which allows us to find $\sin\theta$ and $\cos\theta$.

We use the substitution $x=\tan u$, $\mathrm{d}x=\sec^2 u~\mathrm{d}u$.

$$\begin{aligned} \int\dfrac{\sqrt{1+x^2}^3}{x}\mathrm{d}x&=\int\frac{\sqrt{1+\tan^2 u}^3}{\tan u}\sec^2 u~\mathrm{d}u \\&=\int\frac{\sec^3u}{\tan u}\sec^2 u~\mathrm{d}u \\&=\int\frac{(\sec^2u)^2}{\tan u}\sec u~\mathrm{d}u \\&=\int\frac{(\tan^2u+1)^2}{\tan u}\sec u~\mathrm{d}u \\&=\int\frac{\tan^4u+2\tan^2 u + 1}{\tan u}\sec u~\mathrm{d}u \\&=\int \tan^3 u \sec u ~\mathrm{d}u + \int 2\sec u \tan u~\mathrm{d}u + \int \frac{\sec u}{\tan u}~\mathrm{d}u \\ \text{For the first integral, we use the substitution $w=\sec u$. The second is the antiderivative of $2\sec u$. The third we simplify as $\frac{\sec u}{\tan u} = \frac{1}{\cos u}\cdot \frac{\cos u}{\sin u} = \csc u$~.} \\ &=\int\bigg((\sec^2 u -1)\sec u \tan u\bigg)~\mathrm{d}u + 2\sec u + \log|\cot u - \csc u |+C\\ &=\int (w^2-1)~\mathrm{d}w+2\sec u + \log|\cot u - \csc u|+C\\ &=\frac{1}{3}w^3-w+2\sec u + \log|\cot u - \csc u|+C\\ &=\frac{1}{3}\sec^3u-\sec u+2\sec u + \log|\cot u - \csc u|+C \\ \smash{  } \qquad &=\frac{1}{3}\sec^3u+\sec u + \log|\cot u - \csc u|+C \\ \text{We began with the substitution $x=\tan u$. Then $\cot u = \frac{1}{x}$. To find $\csc u$ and $\sec u$, we draw a right triangle with angle $u$, opposite side $x$, and adjacent side $1$. The Pythagorean Theorem gives us the hypotenuse.} \\ &=\frac{1}{3}\sqrt{1+x^2}^3+\sqrt{1+x^2}+\log\left| \frac{1}{x}-\frac{\sqrt{1+x^2}}{x} \right|+C \\&=\frac{1}{3}\sqrt{1+x^2}(4+x^2)+\log\left|\frac{1-\sqrt{1+x^2}}{x} \right|+C \end{aligned}$$