적분 기법 전체 문제

The derivative of the denominator shows up in the numerator, only differing by a constant, so we perform a substitution. Specifically, substitute $u=x^2-3$, $\mathrm{d}u=2x\,\mathrm{d}x$. This gives

$$\begin{aligned} \int\frac{x}{x^2-3}\mathrm{d}x =\int\frac{\mathrm{d}u/2}{u} =\frac{1}{2}\log|u| + C =\frac{1}{2}\log\big|x^2-3\big| + C \end{aligned}$$

(a) Although a quadratic under a square root often suggests trigonometric substitution, in this case we have an easier substitution. Specifically, let $y=9+x^2$. Then $\mathrm{d}y=2x\mathrm{d}x,\ x\mathrm{d}x=\frac{\mathrm{d}y}{2},\ y(0)=9,$ and $y(4)=25$.

$$\begin{aligned} \int_0^4\frac{x}{\sqrt{9+x^2}}\,\mathrm{d}x &=\int_9^{25}\frac{1}{\sqrt{y}}\,\frac{\mathrm{d}y}{2} =\frac{1}{2}\cdot\frac{\sqrt{y}}{1/2}\Big|_9^{25} =5-3= 2 \end{aligned}$$

(b) The power of cosine is odd, so we can reserve one cosine for the differential and change the rest to sines. Substituting $y=\sin x,$ $\mathrm{d}y=\cos x$, $\mathrm{d}x,$ $y(0)=0,$ $y(\pi/2)=1,$ $\cos^2x=1-y^2$:

$$\begin{aligned} \int_0^{\pi/2}\cos^3x\ \sin^2x\,\mathrm{d}x &=\int_0^{\pi/2}\cos^2x\ \sin^2x\ \cos x\,\mathrm{d}x =\int_0^1\!(1-y^2)y^2\,\mathrm{d}y =\int_0^1(y^2-y^4)\,\mathrm{d}y \\ &=\bigg[\frac{y^3}{3}-\frac{y^5}{5}\bigg]_0^1 =\frac{1}{3}-\frac{1}{5} \\ &= \frac{2}{15} \end{aligned}$$

(c) The integrand is the product of two different kinds of functions, with no obvious substitution or simplification. If we differentiate $\log x$, it will match better with the polynomial nature of the rest of the integrand. So, integrate by parts with $u(x)=\log x$ and $\mathrm{d}v=x^3\,\mathrm{d}x$, then $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=x^4/4$.

$$\begin{aligned} \int_1^{e}x^3\log x\,\mathrm{d}x &=\frac{x^4}{4}\log x\bigg|_1^e -\int_1^{e}\frac{x^4}{4}\cdot\frac{1}{x}\,\mathrm{d}x =\frac{e^4}{4}-\int_1^{e}\frac{x^3}{4}\,\mathrm{d}x =\frac{e^4}{4}-\frac{x^4}{16}\bigg|_1^e \\ &=\frac{3e^4}{16}+\frac{1}{16} \end{aligned}$$

(a) Integrate by parts with $u=x$ and $\mathrm{d}v=\sin x\,\mathrm{d}x$ so that $\mathrm{d}u=\mathrm{d}x$ and $v=-\cos x$.

$$\begin{aligned} \int x\sin x\,\mathrm{d}x&= -x\cos x-\int (-\cos x)\,\mathrm{d}x =-x\cos x+\sin x+C\\ \text{So, }\quad\int_0^{\pi/2} x\sin x\,\mathrm{d}x&=\Big[-x\cos x+\sin x\Big]_0^{\pi/2} =1 \end{aligned}$$

(b) The power of cosine is odd, so we can reserve one cosine for $\mathrm{d}u$ and change the rest into sines. Make the substitution $u=\sin x$, $\mathrm{d}u=\cos x\,\mathrm{d}x$.

$$\begin{aligned} \int_0^{\pi/2} \cos^5 x\,\mathrm{d}x &=\int_0^{\pi/2} \big(1-\sin^2x\big)^2\cos x\,\mathrm{d}x =\int_0^1\big(1-u^2\big)^2\,\mathrm{d}u =\int_0^1\big(1-2u^2+u^4\big)\,\mathrm{d}u\\ &=\left[u-\frac{2}{3}u^3+\frac{1}{5}u^5\right]_0^1 =1-\frac{2}{3}+\frac{1}{5} =\frac{8}{15} \end{aligned}$$

(a) $\int_0^3\sqrt{9-x^2}\,\mathrm{d}x$ is the area of the portion of the disk $x^2+y^2\le 9$ that lies in the first quadrant. It is $\frac{1}{4}\pi 3^3={\frac{9}{4}\pi}\,$. Alternatively, you could also evaluate this integral using the substitution $x=3\sin y$, $\mathrm{d}x=3\cos y\,\mathrm{d}y$.

$$\begin{aligned} \int_0^3\sqrt{9-x^2}\,\mathrm{d}x &=\int_0^{\pi/2}\sqrt{9-9\sin^2y}\ (3\cos y)\,\mathrm{d}y =9\int_0^{\pi/2}\cos^2 y\,\mathrm{d}y \\ &=\frac{9}{2}\int_0^{\pi/2}[1+\cos(2y)]\,\mathrm{d}y =\frac{9}{2}{\Big[y+\frac{\sin(2y)}{2}\Big]}_0^{\pi/2} \\ &=\frac{9}{4}\pi \end{aligned}$$

(b) It's not immediately obvious what to do with this one, but remember we found $\int \log x \mathrm{d}x$ using integration by parts with $u=\log x$ and $\mathrm{d}v=\mathrm{d}x$. Let's hope a similar trick works here. Integrate by parts, using $u=\log(1+x^2)$ and $\mathrm{d}v=\mathrm{d}x$, so that $\mathrm{d}u=\frac{2x}{1+x^2}\,\mathrm{d}x$, $v=x$.

$$\begin{aligned} \int_0^1\log(1+x^2)\,\mathrm{d}x &=\Big[x\log(1+x^2)\Big]_0^1-\int_0^1 x\frac{2x}{1+x^2}\,\mathrm{d}x =\log 2-2\int_0^1 \frac{x^2}{1+x^2}\,\mathrm{d}x \\ &=\log 2-2\int_0^1\Big(1- \frac{1}{1+x^2}\Big)\,\mathrm{d}x =\log 2-2\big[x-\arctan x\big]_0^1 \\ &=\log 2-2+\frac{\pi}{2}\approx 0.264 \end{aligned}$$

(c) The integrand is a rational function with no obvious substitution, so we use partial fractions.

$$\begin{aligned} \frac{x}{(x-1)^2(x-2)} &=\frac{a}{(x-1)^2}+\frac{b}{x-1}+\frac{c}{x-2} =\frac{a(x-2)+b(x-1)(x-2)+c(x-1)^2}{(x-1)^2(x-2)} \\ \\ \text{Multiply by the denominator.} \\ x&=a(x-2)+b(x-1)(x-2)+c(x-1)^2 \end{aligned}$$

Setting $x=1$ gives $a=-1$. Setting $x=2$ gives $c=2$. Substituting in $a=-1$ and $c=2$ gives

$$\begin{aligned} b(x-1)(x-2)&=x+(x-2)-2(x-1)^2=-2x^2+6x-4=-2(x-1)(x-2) \\ \implies b&=-2 \end{aligned}$$

Hence

$$\begin{aligned} \int_3^\infty\frac{x}{(x-1)^2(x-2)}\,\mathrm{d}x &=\lim_{M\rightarrow\infty} \int_3^M\left(-\frac{1}{(x-1)^2}-\frac{2}{x-1}+\frac{2}{x-2}\right)\,\mathrm{d}x \\ &=\lim_{M\rightarrow\infty} \left[\frac{1}{x-1}-2\log|x-1|+2\log|x-2|\right]_3^M\\ &=\lim_{M\rightarrow\infty} \bigg[\frac{1}{x-1}+2\log\bigg|\frac{x-2}{x-1}\bigg|\bigg]_3^M\\ &=\lim_{M\rightarrow\infty} \bigg[\frac{1}{M-1}+2\log\bigg|\frac{M-2}{M-1}\bigg|\bigg] -\bigg[\frac{1}{3-1}+2\log\bigg|\frac{3-2}{3-1}\bigg|\bigg] \\ &=2\log 2-\frac{1}{2}\approx0.886 \end{aligned}$$

since

$$\begin{aligned} \lim_{M\rightarrow\infty}\log\frac{M-2}{M-1} &=\lim_{M\rightarrow\infty}\log\frac{1-2/M}{1-1/M} =\log 1=0\\ \text{and}\qquad \log\frac{1}{2}&=-\log2 \end{aligned}$$

(a) It doesn't matter to us right now that the arguments of sine and cosine are $2x$ rather than $x$. This is still the integral of powers of products of sines and cosines. Since cosine has an odd power, we make the substitution $u=\sin (2x)$, $\mathrm{d}u=2\cos(2 x)\,\mathrm{d}x$.

$$\begin{aligned} \int_0^{\pi\over 4}\sin^2(2x)\cos^3(2x)\ \mathrm{d}x &=\int_0^{\pi\over 4} \sin^2(2x)\big[1-\sin^2(2x)\big]\cos(2 x)\ \mathrm{d}x =\frac{1}{2}\int_0^1 u^2\big[1-u^2\big]\ \mathrm{d}u\\ &=\frac{1}{2}\int_0^1\big(u^2-u^4\big)\ \mathrm{d}u =\frac{1}{2}{\Big[\frac{1}{3}u^3-\frac{1}{5}u^5\Big]}_0^1 =\frac{1}{15} \end{aligned}$$

(b) Make the substitution $x=3\tan t$, $\mathrm{d}x=3\sec^2t\,\mathrm{d}t$ and use the trig identity $9+9\tan^2 t = 9\sec^2 t$.

$$\begin{aligned} \int\big(9+x^2\big)^{-{3\over 2}}\ \mathrm{d}x &=\int\big(9+9\tan^2t\big)^{-{3\over 2}}\ 3\sec^2t\,\mathrm{d}t =\int\big(3\sec t\big)^{-3}\ 3\sec^2t\,\mathrm{d}t \\ &=\frac{1}{9}\int\cos t\ \mathrm{d}t =\frac{1}{9}\sin t+C =\frac{1}{9}\frac{x}{\sqrt{x^2+9}}+C \end{aligned}$$

To convert back to $x$, in the last step, we used the triangle below, which is rigged to have $\tan t =\frac{x}{3}$.

(c) Seeing a rational function with no obvious substitutions, we use partial fractions.

$$\begin{aligned} \frac{1}{(x-1)(x^2+1)} &=\frac{a}{x-1}+\frac{bx+c}{x^2+1} =\frac{a(x^2+1)+(bx+c)(x-1)}{(x-1)(x^2+1)} \\ \text{Multiply by the original denominator.} \\ 1&=a(x^2+1)+(bx+c)(x-1)\tag{$*$} \end{aligned}$$

Setting $x=1$ gives $2a=1$ or $a=\tfrac{1}{2}$. Substituting in $a=\tfrac{1}{2}$ in $(*)$ gives

$$\begin{aligned} &\frac{1}{2}(x^2+1)+(bx+c)(x-1)=1\\ \iff\qquad& (bx+c)(x-1)=\frac{1}{2}(1-x^2)=-\frac{1}{2}(x-1)(x+1)\\ \iff\qquad& (bx+c)=-\frac{1}{2}(x+1)\\ \iff\qquad& b=c=-\frac{1}{2}\\ \end{aligned}$$

So,

$$\begin{aligned} \int\frac{\mathrm{d}x}{(x-1)(x^2+1)} &=\int\Big[\frac{1/2}{x-1}-\frac{\frac{1}{2}(x+1)}{x^2+1}\Big]\,\mathrm{d}x \\ \text{To antidifferentiate the second piece, we split it into two integrals: one that can be handled with the substitution $u=x^2+1$, and another that looks like the derivative of arctangent.} \\ &=\int\Big(\frac{1/2}{x-1}-\frac{x/2}{x^2+1} -\frac{1/2}{x^2+1}\Big)\,\mathrm{d}x\\ &=\int\Big(\frac{1/2}{x-1}-\frac{1}{4}\cdot\frac{2x}{x^2+1} -\frac{1/2}{x^2+1}\Big)\,\mathrm{d}x\\ &=\frac{1}{2}\log|x-1|-\frac{1}{4}\log(x^2+1)-\frac{1}{2}\arctan x+C \end{aligned}$$

(d) We know the derivative of arctangent, and it would integrate nicely if multiplied to the antiderivative of $x$. So, we integrate by parts with $u=\arctan x$ and $\mathrm{d}v=x\,\mathrm{d}x$ so that $\mathrm{d}u=\frac{1}{1+x^2}\mathrm{d}x$ and $v=\frac{1}{2} x^2$. Then

$$\begin{aligned} \int x\arctan x\,\mathrm{d}x&= \frac{1}{2} x^2\arctan x -\frac{1}{2}\int \frac{x^2}{1+x^2}\,\mathrm{d}x \\ &= \frac{1}{2} x^2\arctan x-\frac{1}{2}\int \frac{1+x^2-1}{1+x^2}\,\mathrm{d}x\\ &= \frac{1}{2} x^2\arctan x-\frac{1}{2}\int \left(1-\frac{1}{1+x^2}\right)\,\mathrm{d}x\\ &=\frac{1}{2}\big[ x^2\arctan x -x +\arctan x\big]+C \end{aligned}$$