치환적분법

We use the substitution $u=4-x^2$, for which $\mathrm{d}u=-2x\,\mathrm{d}x$ :

$$\begin{aligned} \int \frac{x^2-4}{\sqrt{4-x^2}}\,x\,\mathrm{d}x &=\int \frac{1}{2}\cdot\frac{ 4-x^2}{\sqrt{ 4-x^2}} ({-}2x)\,\mathrm{d}x \\ &=\frac{1}{2} \int \frac{{red}u}{\sqrt{{red}u}}\, \mathrm{d}u \\ &=\frac{1}{2}\int \sqrt{u}\,\mathrm{d}u \\ &=\frac{1}{2}\frac{u^{3/2}}{3/2}+C\\ &=\frac{1}{3}(4-x^2)^{3/2}+C \end{aligned}$$

• Solution 1: If we let $u=\sqrt{\log x}$, then $\mathrm{d}u=\dfrac{1}{2x\sqrt{\log x}}\ \mathrm{d}x$, and:

$$\begin{aligned} \int \dfrac{e^{ \sqrt{\log x}}}{ 2x\sqrt{\log x}}\ \mathrm{d}x&=\int e^{{red}u}\ \mathrm{d}u=e^u+C=e^{\sqrt{\log x}}+C \end{aligned}$$

• Solution 2: In Solution 1, we made a pretty slick choice. We might have tried to work with something a little less convenient. For example, it's not unnatural to think that $u=\log x$, $\mathrm{d}u=\dfrac{1}{x}\ \mathrm{d}x$ would be a good choice. In that case:

$$\begin{aligned} \int \dfrac{e^{\sqrt{ \log x}}}{2 x\sqrt{ \log x}}\ \mathrm{d}x&= \int \frac{e^{\sqrt{{red}u}}}{2\sqrt{{red}u}} \mathrm{d}u \\ \text{Now, we should be able to see that $ w=\sqrt{u}$, $ \mathrm{d}w = \dfrac{1}{2\sqrt{u}}\ \mathrm{d}u$ is a good choice:} \\ \int\frac{e^{ \sqrt{u}}}{ 2\sqrt{u}}\ \mathrm{d}u &=\int e^{{orange}w}\ \mathrm{d}w\\ &=e^{\sqrt{u}}+C\\ &=e^{\sqrt{\log x}}+C \end{aligned}$$

(a) We let $u = \cos(x)$, so $du = -\sin(x) \ dx$. This change of variable transforms the integral into one we can integrate as follows: $$\begin{aligned}\int \tan(x) \, dx &= \int \frac{\sin(x)}{\cos(x)} \ dx \\ &= -\int \frac{1}{u} \ du \\ &= -\ln(|u|) + C \\ &= -\ln(|\cos(x)| + C \\ &= \ln\left(|\cos(x)|^{-1}\right) + C \\ &= \ln\left(|\sec(x)|\right) + C\end{aligned}\,.$$ Note that we used the property of the logarithm that $p \ln (z) = \ln(z^p)$ as well as the fact that $\frac{1}{\cos(x)} = \sec(x)$ in order to simplify the result. (b) We let $u = \sin(x)$, so $du = \cos(x) \ dx$. This change of variable transforms the integral into one we can integrate as follows: $$\begin{aligned}\int \cot(x) \, dx &= \int \frac{\cos(x)}{\sin(x)} \ dx \\ &= \int \frac{1}{u} \ du \\ &= \ln(|u|) + C \\ &= \ln(|\sin(x)| + C \\ &= -\ln\left(|\sin(x)|^{-1}\right) + C \\ &= -\ln\left(|\csc(x)|\right) + C\end{aligned}\,.$$ Note that here we used the fact that $\ln(z) = (-1)(-1)\ln(z) = (-1) \ln(z^{-1})$ in order to simplify the result. (c) We let $u = \sec(x) + \tan(x)$, so it follows that $du = \left(\sec(x)\tan(x)+\sec^2(x)\right) \ dx$. This change of variable shows that the basic structure of the integrand is simple and can be integrated as follows: $$\begin{aligned}\int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} \, dx &= \int \frac{1}{u} \ du \\ &= \ln(|u|) + C \\ &= \ln\left(|\sec(x) + \tan(x)|\right) + C\end{aligned}\,.$$ (d) The numerator of $\frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)}$ can be rewritten by removing the common factor of $\sec(x)$ to obtain $$\sec(x)\left(\sec(x) + \tan(x)\right)\,.$$ The integrand can then be simplified to $$\frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} = \frac{\sec(x)\left(\sec(x) + \tan(x)\right)}{\sec(x) + \tan(x)} = \sec(x)\,.$$ (e) Using the results of (c) and (d), we see that $$\int \sec(x) \, dx = \int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} \, dx = \ln\left(|\sec(x) + \tan(x)|\right) + C\,.$$ (f) Following (c)-(e) we can evaluate $int \csc(x) \, dx$ as follows. First, rewrite $\csc(x)$ by multiplying by $1 = \frac{\csc(x) + \cot(x)}{\csc(x) + \cot(x)}$, so that $$\int \csc(x) \, dx = \int \frac{\csc(x)\left(\csc(x) + \cot(x)\right)}{\csc(x) + \cot(x)} \, dx = \int \frac{\csc^2(x) + \csc(x) \cot(x)}{\csc(x) + \cot(x)} \, dx\,.$$ Now we evaluate the last integral in the preceding equation with the substitution $u = \csc(x) + \cot(x)$. Then $du = -\left(\csc(x)\cot(x)+\csc^2(x)\right) \ dx$. This change of variable transforms the integral so that we can integrate: $$\begin{aligned}\int \frac{\csc^2(x) + \csc(x) \cot(x)}{\csc(x) + \cot(x)} \, dx &= - \int \frac{1}{u} \ dx \\ &= -\ln(|u|) + C \\ &= -\ln\left(|\csc(x) + \cot(x)|\right) + C\end{aligned}\,.$$ Connecting the equalities we have established in our work, we have shown that $$\int \csc(x) \, dx = -\ln\left(|\csc(x) + \cot(x)|\right) + C\,.$$

(a) Letting $u = x-1$, we find that $x = u+1$ and thus $dx = du$. (b) Using the change of variables in (a), $$\int x \sqrt{x-1} \, dx = \int (u+1) \sqrt{u} \, du\,.$$ (c) Using the fact that $\sqrt{u} = u^{1/2}$ and distributing, we can evaluate the integral as follows: $$\begin{aligned}\int x \sqrt{x-1} \, dx &= \int (u+1) u^{1/2} \, du \\ &= \int \left(u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\,.$$ (d) First, for $\int x^2 \sqrt{x-1} \, dx$, we let $u = x-1$ so that $x = u+1$ and $dx = du$. From there, our work is very similar to what we did in (b): $$\begin{aligned}\int x^2 \sqrt{x-1} \, dx &= \int (u+1)^2 u^{1/2} \, du \\ &= \int \left(u^2 + 2u + 1 \right) u^{1/2} \, du \\ &= \int \left(u^{5/2} + 2u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{7} u^{7/2} + \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{7} (x-1)^{7/2} + \frac{4}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\,.$$ For $\int x \sqrt{x^2 - 1} \, dx$, despite the fact that the integral looks remarkably similar, the situation is quite different. We observe that with $u = x^2 - 1$, it follows $du = 2x \, dx$, so we have a standard function-derivative pair. Noting that $x \, dx = \frac{1}{2} \, du$, it follows that $$\begin{aligned}\int x \sqrt{x^2 - 1} \, dx &= \int \sqrt{u} \cdot \frac{1}{2} \, du \\ &= \frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C \\ &= \frac{1}{3} u^{3/2} + C \\ &= \frac{1}{3} (x^2-1)^{3/2} + C\end{aligned}\,.$$ In the situation where we have $(x-a)$ under a square root and a polynomial outside the square root, the substitution $u = x-a$ will allow us to convert the integrand into an expression we can integrate by changing the form of the radical from $\sqrt{x-a}$ to $\sqrt{u}$. From there we are able to distribute and integrate an expression that is essentially a sum of (fractional) powers of $u$. In the situation where a polynomial of degree higher than $1$ (such as $x^2 - 1$) is under the radical, we likely have to be fortunate enough to have the derivative of said polynomial (up to a constant) outside the radical in order to be able to integrate, as we were in the last example.

(a) If we let $u = \sin(x)$, then $du = \cos(x) \, dx$. Since there is no cosine function present in the integrand, we don't have the needed function-derivative pair present in order to transform the integral into one we can evaluate. (b) Using the Fundamental Trigonometric Identity, we observe that $\sin^2(x) = 1-\cos^2(x)$, and thus $$\sin^3(x) = \sin(x) \cdot \sin^2(x) = \sin(x) (1-\cos^2(x))\,.$$ (c) By our work in (b), we see that $\int \sin^3(x) \, dx = \int \sin(x) (1-\cos^2(x)) \, dx$, and in this integral we have the function-derivative pair that corresponds to $u = \cos(x)$ and $du = -\sin(x) \, dx$. (d) Using this substitution and evaluating the resulting integral, $$\begin{aligned}\int \sin^3(x) \, dx &= \int \sin(x) (1-\cos^2(x)) \, dx \\ &= \int (-1)(1-u^2) \, du \\ &= \int (u^2-1) \, du \\ &= \frac{1}{3}u^3 - u + C \\ &= \frac{1}{3}\cos^3(x) - \cos(x) + C\end{aligned}\,.$$ (e) Here we observe that $\cos^3(x) = \cos(x) \cdot \cos^2(x) = \cos(x) (1-\sin^2(x))$, so letting $u = \sin(x)$ and thus $du = \cos(x) \, dx$, we have $$\begin{aligned}\int \cos^3(x) \, dx &= \int \cos(x) (1-\sin^2(x)) \, dx \\ &= \int (1-u^2) \, du \\ &= u - \frac{1}{3}u^3 + C \\ &= \sin(x) - \frac{1}{3}\sin^3(x) + C\end{aligned}\,.$$

(a) In the following plot, we see that the model is reasonable because it appears to be periodic and repeats every $24$ hours, as well as that the rate of consumption seems to peak at the times of day where people are most active in their homes: around $7$ a.m. and around $5$ p.m. (b) The meaning of $\int_0^{24} r(t) \, dt$ is the total power consumed in $24$ hours, measured in megawatt-hours. This is because $r(t)$ is measured in megawatts and $\triangle t$ in hours, and thus in the Riemann sum that corresponds to the definite integral, the terms in the sum of the form $r(t) \cdot \triangle t$ have units “megawatt-hours”. Moreover, by integrating the rate of power consumption we get the total change in power consumed. (c) In a typical day, we know that Mathland consumes $\int_0^{24} r(t) \, dt$ megawatt-hours of power, and thus we seek to evaluate the integral $$\int_0^{24} r(t) \, dt = \int_0^{24} \left[ 4 + \sin(0.263t + 4.7) + \cos(0.526t+9.4) \right] \, dt\,.$$ To do so, we use the subtitutions $u = 0.263t + 4.7$ (so $du = 0.263 \, dt$) and $w = 0.526t + 9.4$ (so $dw = 0.526 \, dt$) to help us antidifferentiate the two trigonometric expressions, and thus we find that $$\int \sin(0.263t + 4.7) \, dt = \frac{1}{0.263} \int \sin(u) \, du = -\frac{1}{0.263} \cos(u) = -\frac{1}{0.263} \cos(0.263t + 4.7)$$ and $$\int \cos(0.526t + 9.4) \, dt = \frac{1}{0.526} \int \cos(w) \, dw = \frac{1}{0.526} \sin(w) = \frac{1}{0.526} \sin(0.526t + 9.4)\,.$$ Putting this all together and using the Fundamental Theorem of Calculus, we have $$\begin{aligned}\int_0^{24} r(t) \, dt &= \left. 4t -\frac{1}{0.263} \cos(0.263t + 4.7) + \frac{1}{0.526} \sin(0.526t + 9.4) \right|_0^{24} \\ &= \left[ 4 \cdot 24 -\frac{1}{0.263} \cos(0.263 \cdot 24 + 4.7) + \frac{1}{0.526} \sin(0.526 \cdot 24 + 9.4) \right] - \left[ -\frac{1}{0.263} \cos(4.7) + \frac{1}{0.526} \sin(9.4) \right] \\ &\approx 95.7809\end{aligned}$$ megawatt-hours of power used in $24$ hours. (d) We know that the average value of the rate function $r$ on $[0,24]$ is $$\displaystyle r_{\operatorname{AVG} [0,24]} = \frac{1}{24-0} \int_{0}^{24} r(t) \, dt\,.$$ Thus, by our work in (c), we have that $$\displaystyle r_{\operatorname{AVG} [0,24]} \approx \frac{1}{24} \cdot 95.7809 = 3.99087$$ which is measured in megawatts.