치환적분법

(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 $$\int \csc(x) \, dx = -\ln\left(|\csc(x) + \cot(x)|\right) + C\,.$$