\(\displaystyle\fbox{56}***\qquad \int \frac{1}{\sin x}dx \) \begin{align*} \frac{1}{\sin x}&=\frac{\sin x}{\sin^2x}\\ &=\frac{\sin x}{1-\cos^2x}\\ &=\frac{1}{t^2-1}\\ &=\frac{1}{(t+1)(t-1)}\\ =\frac{1}{2}&\left(\frac{1}{t-1}-\frac{1}{t+1}\right) \end{align*} ここで \begin{align*} t&=\cos x\\ dt&=-\sin xdx\\ \left|\frac{\cos x-1}{\cos x+1}\right|&=\frac{1-\cos x}{\cos x+1}\\ \end{align*} と置換すると, \begin{align*} \int \frac{1}{\sin x}dx&=\frac{1}{2}\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right) dt\\ &=\frac{1}{2}\log |\frac{t-1}{t+1}|+C\\ &=\frac{1}{2}\log \frac{1-\cos x}{1+\cos x}+C\qquad(答) \end{align*}

\(\displaystyle \fbox{57}**\qquad\int \tan^3x dx\) \begin{align*} %Q57 \int \tan^3xdx&=\int \tan x \cdot \tan^2x dx\\ &=\int \tan x \cdot (\frac{1}{\cos^2x}-1)dx\\ &=\int \tan x \cdot \frac{1}{\cos^2x} + \frac{-\sin x}{\cos x}dx\\ &=\int \tan x \cdot (\tan x)' + \frac{(\cos x)'}{\cos x}dx\\ &=\frac{1}{2}\tan^2x + \log |\cos x| + C\qquad(答)\\ \end{align*}

\(\displaystyle \fbox{58}****\qquad\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin x+\cos x+1}dx\\\) \(t=\tan \frac{x}{2}\)と置換する \begin{flalign*} \sin x&=\frac{\sin x}{1}\\ &=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}\\ &=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\\ &=\frac{2t}{1+t^2}\\ \cos x&=\frac{\cos x}{1}\\ &=\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}\\ &=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\\ &=\frac{1-t^2}{1+t^2} \end{flalign*} \begin{flalign*} \tan x&=\frac{\sin x}{\cos x}\\ &=\frac{2t}{1+t^2}\cdot\frac{1+t^2}{1-t^2}\\ &=\frac{2t}{1-t^2}\\ \\ t&=\tan^2\frac{x}{2}を微分して,\\ \frac{dt}{dx}&=\frac{1}{\cos^2\frac{x}{2}}\cdot\frac{1}{2}\\ &=\frac{1}{2}\left(1+\tan^2\frac{x}{2}\right)\\ &=\frac{1+t^2}{2} \end{flalign*} ここで \begin{align*} dx=\frac{2}{1+t^2}dt\quad\\ \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & \frac{\pi}{2} \\ \hline t & 0 & \rightarrow & 1 \\ \hline \end{array} と置換している \begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{1}{\sin x+\cos x+1}dx&=\int_{0}^{1}\frac{1}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}+1}\cdot\frac{2}{1+t^2}dt\\ &=\int_{0}^{1}\frac{2}{2t+(1-t^2)+(1+t^2)}dt\\ &=\int_{0}^{1}\frac{1}{1+t}dt\\ &=\left[\log(1+t)\right]^1_0=\log2\qquad(答) \end{align*}

\(\displaystyle \fbox{59}**\qquad\int\sin\left(\log x\right) dx\) \begin{align*} t&=\log x\\ x&=e^t\\ dx&=e^t dt\\ \end{align*} と置換して, \begin{align*} \int\sin\left(\log x\right) dx&=\int\sin t e^t dt\\ &=\frac{1}{2}e^t\left(\sin t-\cos t\right)+C\\ &=\frac{1}{2}x \lbrace\sin\left(\log x\right)-\cos\left(\log x\right)\rbrace+C\quad(答) \end{align*}

\(\displaystyle \fbox{60}**\qquad\int\cos{2x}\cos{4x}dx\) \begin{align*} %Q60 \int\cos{2x}\cos{4x}dx&=\frac{1}{2}\int\left(\cos{6x}+\cos{2x}\right)dx\\ &=\frac{1}{2}\left(\frac{1}{6}\sin6x+\frac{1}{2}\sin2x\right)+C\\ &=\frac{1}{12}\sin6x+\frac{1}{4}\sin2x+C\qquad(答) \end{align*}

\(\displaystyle\fbox{61}*****\quad\int^{\frac{\pi}{2}}_0\frac{(\cos x)^{\sqrt{3}}}{(\sin x)^{\sqrt{3}}+(\cos x)^{\sqrt{3}}}dx \) \begin{align*} x&=\frac{\pi}{2}-t\quad\quad\;\\ dx&=-dt \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & \frac{\pi}{2} \\ \hline t & \frac{\pi}{2} & \rightarrow & 0 \\ \hline \end{array} と置換して, \begin{align*} I&=\int^{\frac{\pi}{2}}_0\frac{(\cos x)^{\sqrt{3}}}{(\sin x)^{\sqrt{3}}+(\cos x)^{\sqrt{3}}}dx \\ &=\int^{\frac{\pi}{2}}_0\frac{(\sin t)^{\sqrt{3}}}{(\cos t)^{\sqrt{3}}+(\sin t)^{\sqrt{3}}}dt \\ &=\int^{\frac{\pi}{2}}_0\frac{(\cos x)^{\sqrt{3}}}{(\sin x)^{\sqrt{3}}+(\cos x)^{\sqrt{3}}}dx \\ \therefore2I&=\int^{\frac{\pi}{2}}_0 1dx=\frac{\pi}{2}\\ I&=\frac{\pi}{4}\quad(答) \end{align*}

\(\displaystyle \fbox{62}**\qquad\int\cos{x}\cdot\cos{2x}\cdot\cos{3x}dx\) \begin{align*} %Q62 \cos{x}\cos{2x}\cos{3x}&=\frac{\cos{3x}+\cos{x}}{2}\cdot\cos{3x}\\ &=\frac{1}{2}(\cos^2{3x}+\cos{x}\cos{3x})\\ &=\frac{1}{2}\left(\frac{1+\cos{6x}}{2}+\frac{\cos{4x}+\cos{2x}}{2}\right)\\ &=\frac{1}{4}(1+\cos{2x}+\cos{4x}+\cos{6x})\\ \therefore\int\cos{x}\cos{2x}\cos{3x}dx&=\frac{1}{4}(x+\frac{1}{2}\sin{2x}+\frac{1}{4}\sin{4x}+\frac{1}{6}\sin{6x})+C\\ &=\frac{1}{48}(12x+6\sin{2x}+3\sin{4x}+2\sin{6x})+C\qquad(答) \end{align*}

\(\displaystyle \fbox{63}****\quad\int^{\frac{\pi}{4}}_0\frac{\cos x\sin x}{\cos^4 x+\sin^4 x}dx\) \begin{align*} \tan^2 x&=u\;\\ (\tan^2 x)'dx&=du \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & \frac{\pi}{4} \\ \hline u & 0 & \rightarrow & 1 \\ \hline \end{array} \begin{align*} u&=\tan\theta\quad\quad\;\;\;\\ du&=\frac{d\theta}{\cos^2\theta} \end{align*} \begin{array}{|c||ccc|} %増減表 \hline u & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} と置換して, \begin{align*} \int^{\frac{\pi}{4}}_0\frac{\cos x\sin x}{\cos^4 x+\sin^4 x}dx&=\int^{\frac{\pi}{4}}_0\frac{\frac{1}{\cos^4 x}}{\frac{1}{\cos^4 x}}\frac{\cos x\sin x}{\cos^4 x+\sin^4 x}dx\\ &=\int^{\frac{\pi}{4}}_0\frac{\tan x\cdot\frac{1}{\cos^2 x}}{1+\tan^4 x}dx\\ &=\int^{\frac{\pi}{4}}_0\frac{\frac{1}{2}(\tan^2 x)'}{1+(\tan^2 x)^2}dx\\ &=\int^1_0\frac{1}{2}\frac{1}{1+u^2}du\\ &=\int^{\frac{\pi}{4}}_0\frac{1}{2}\frac{1}{1+\tan^2 \theta}\frac{d\theta}{\cos^2\theta}\\ &=\frac{1}{2}\int^{\frac{\pi}{4}}_0 d\theta=\frac{\pi}{8}\qquad(答) \end{align*}

\(\displaystyle\fbox{64}*****\quad\int_{0}^{\frac{\pi}{4}}\frac{x^2}{(x\sin x+\cos x)^2}dx \) \begin{align*} %Q64 \int_{0}^{\frac{\pi}{4}}\frac{x^2}{(x\sin x+\cos x)^2}dx&=\int_{0}^{\frac{\pi}{4}}\underset{微分}{\underline{(\frac{x}{\cos x})}}\underset{積分}{\underline{\frac{x\cos x}{(x\sin x+\cos x)}}}dx\\ &=\left[-\frac{x}{\cos x}\frac{1}{x\sin x+\cos x}\right]^{\frac{\pi}{4}}_0 +\int_{0}^{\frac{\pi}{4}}\frac{\cancel{\cos x+x\sin x}}{\cos^2x}\frac{dx}{\cancel{x\sin x+\cos x}}\\ &=-\frac{\frac{\pi}{4}}{\frac{1}{\sqrt{2}}}\frac{1}{\frac{\pi}{4}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}+\left[\tan x\right]^{\frac{\pi}{4}}_0\\ &=-\frac{\pi}{4}\frac{1}{\frac{\pi}{4}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}+1\\ &=-\frac{\pi}{\pi+4}+1=\frac{4-\pi}{4+\pi}\qquad(答) \end{align*}

\(\displaystyle \fbox{65}*****\quad\int^1_{-1}\frac{\sin^2(\pi x)}{1+e^x}dx\) \begin{align*} %Q65 I&=\int^1_{-1}\frac{\sin^2(\pi x)}{1+e^x}dx\\ 2I&=\int^1_{-1}\left\lbrace\frac{\sin^2(\pi x)}{1+e^x}\frac{\sin^2(\pi x)}{1+e^{-x}}\right\rbrace dx\\ &=\int^1_{-1}\frac{\cancel{(1+e^x)}\sin^2(\pi x)}{\cancel{1+e^x}} dx\\ &=\int^1_{-1}\sin^2(\pi x)dx\\ &=\int^1_{-1}\frac{1-\cos(2\pi x)}{2}dx\\ &=\left[\frac{1}{2}x-\frac{1}{4\pi}\sin(2\pi x)\right]^1_{-1}=1\\ \therefore I&=\frac{1}{2}\qquad(答) \end{align*}

\(\displaystyle \fbox{66}**\qquad\int^{\pi}_0 e^{2x}\sin xdx\) \begin{align*} %Q66 I&=\int^{\pi}_0 e^{2x}\sin xdx\\ &=\left[\frac{1}{2}e^{2x}\sin x\right]^{\pi}_0-\frac{1}{2}\int^{\pi}_0e^{2x}\cos xdx\\ &=-\frac{1}{2}\left\lbrace\left[\frac{1}{2}e^{2x}\cos x\right]^{\pi}_0+\frac{1}{2}\underline{\int^{\pi}_0e^{2x}\sin xdx}\right\rbrace\\ &=-\frac{1}{2}\left(-\frac{1}{2}e^{2\pi}-\frac{1}{2}+\frac{1}{2}I\right)\\ %\frac{5}{4}I&=\frac{1}{4}\left(e^{2\pi}+1\right)\\ I&=\frac{1}{5}\left(e^{2\pi}+1\right)\qquad(答) \end{align*}

\(\displaystyle \fbox{67}*****\quad\int^{\pi}_0\frac{x\sin x}{3+\cos{2x}}dx\) \begin{align*} x&=\pi-t\quad\quad\;\;\\ dx&=-dt \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & \pi \\ \hline t & \pi & \rightarrow & 0 \\ \hline \end{array} \begin{align*} t&=\cos x\quad\qquad\;\\ dt&=-\sin x dx \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & \pi \\ \hline t & 1 & \rightarrow & -1 \\ \hline \end{array} \begin{align*} t&=\tan\theta\qquad\quad\;\;\\ dt&=\frac{1}{\cos^2\theta} d\theta \end{align*} \begin{array}{|c||ccc|} %増減表 \hline t & -1 & \rightarrow & 1 \\ \hline \theta & -\frac{\pi}{4} & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} と置換して, \begin{align*} -証&明-\\ I=\int^{\pi}_0xf(\sin x)dx&=\int^0_{\pi}(\pi-t)f(\sin(\pi-t))(-dt)\\ &=\int^{\pi}_0(\pi-t)f(\sin t)dt\\ &=\pi\int^{\pi}_0f(\sin t)dt-\int^{\pi}_0tf(\sin t)dt\\ これ&を用いて\\ \int^{\pi}_0&\frac{x\sin x}{3+\cos{2x}}dx\\ &=\frac{\pi}{2}\int^{\pi}_0\frac{\sin x}{3+\cos{2x}}dx\\ &=\frac{\pi}{4}\int^{\pi}_0\frac{\sin x}{1+\cos^2{x}}dx\\ &=\frac{\pi}{4}\int^{\pi}_0\frac{1}{1+t^2}dt\\ &=\frac{\pi}{4}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}d\theta=\frac{\pi^2}{8}\qquad(答)\\ \end{align*}

\(\displaystyle \fbox{68}***\qquad\int(x^2-2x)\cos{2x}dx\) \begin{align*} %Q68 \int(x^2-2x)\cos{2x}dx&=(x^2-2x)\frac{1}{2}\sin{2x}-\int(2x-2)\cdot\frac{1}{2}\sin{2x}dx\\ &=(x^2-2x)\frac{1}{2}\sin{2x}-(2x-2)(-\frac{1}{4}\cos{2x})+\int 2(-\frac{1}{4}\cos{2x})dx\\ &=(x^2-2x)\frac{1}{2}\sin{2x}-(2x-2)(-\frac{1}{4}\cos{2x})-\frac{1}{2}\cdot\frac{1}{2}\sin{2x}+C\\ =&\frac{1}{4}\left\lbrace(2x^2-4x-1)\sin{2x}+2(x-1)\cos{2x}\right\rbrace+C\;(答) \end{align*}

\(\displaystyle \fbox{69}***\qquad\int^{\frac{\pi}{2}}_0\sin^8xdx\) \begin{align*} %Q69 I_8&=\int^{\frac{\pi}{2}}_0\sin^8xdx\\ I_n&=\int^{\frac{\pi}{2}}_0\sin^n xdxとおくと,\\ &=\cancel{\left[-\cos x\sin^{n-1}x\right]^{\frac{\pi}{2}}_0}+(n-1)\int^{\frac{\pi}{2}}_0\cos^2 x\sin^{n-2}xdx\\ &=(n-1)\int^{\frac{\pi}{2}}_0(1-\sin^2x)\sin^{n-2}xdx\\ &=(n-1)\left\lbrace I_{n-2}-I_n \right\rbrace\\ \therefore I_n&=\frac{n-1}{n}I_{n-2}\\ I_8&=\frac{7}{8}I_6=\frac{7}{8}\cdot\frac{5}{6}I_4=\cdots=\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}I_0 となるので,\\ I_0&=\int^{\frac{\pi}{2}}_0 dx=\frac{\pi}{2}より,\\ I_8&=\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}=\frac{35}{256}\pi\qquad(答) \end{align*}

\(\displaystyle \fbox{70}****\qquad\int^{\frac{\pi}{4}}_0\tan^8xdx\) \begin{align*} %Q70 J_8&=\int^{\frac{\pi}{4}}_0\tan^8xdx\\ J_n&=\int^{\frac{\pi}{4}}_0 (\frac{1}{\cos^2x}-1)\tan^{n-2}x dx\\ &=\int^{\frac{\pi}{4}}_0 (\tan x)'\tan^{n-2}x dx-J_{n-2}\\ &=\left[\frac{\tan^{n-1}x}{n-1}\right]^{\frac{\pi}{4}}_0-J_{n-2}\\ \therefore J_n&=-J_{n-2}+\frac{1}{n-1}\\ J_8&=-J_6+\frac{1}{7}=-(-J_4+\frac{1}{5})+\frac{1}{7}\\ &=J_4-\frac{1}{5}+\frac{1}{7}=-J_2+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}\\ &=-(-J_0+\frac{1}{1})+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}=J_0-1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}\\ J_0&=\int^{\frac{\pi}{4}}_0dx=\frac{\pi}{4}と計算して\\ J_8&=\frac{\pi}{4}-\frac{76}{105}\qquad(答) \end{align*}

\(\displaystyle \fbox{71}***\qquad\int^{\frac{\pi}{3}}_0\frac{1}{\cos x}dx\) \begin{align*} %Q71 \int\frac{1}{\cos x}dx&=\int\frac{\cos x}{\cos^2 x}dx\\ &=\int\frac{\cos x}{1-\sin^2x}dx \end{align*} \begin{align*} t&=\sin{x}\quad\quad\;\\ dt&=\cos{x}dx\\ \end{align*} と置換して, \begin{align*} &=\int\frac{1}{1-t^2}dt\\ &=\int\frac{1}{(1+t)(1-t)}dt\\ &=\frac{1}{2}\int\left(\frac{1}{1+t}+\frac{1}{1-t}\right)dt\\ &=\frac{1}{2}\left\lbrace\log|1+t|-\log|1-t|\right\rbrace+C\\ &=\frac{1}{2}\log\left|\frac{1+t}{1-t}\right|+C\\ &=\frac{1}{2}\log\left(\frac{1+\sin x}{1-\sin x}\right)+C \end{align*}

\(\displaystyle \fbox{72}\qquad\int\frac{1}{\cos^3 x}dx\) \begin{align*} %Q72 \sin{x}=&t, \cos{x}dx=dtと置換する\\ \int\frac{1}{\cos^3 x}dx&=\int\frac{\cos{x}}{\cos^4{x}}dx\\ &=\int\frac{\cos{x}}{(1-\sin^2{x})^2}dx\\ &=\int\frac{1}{(1-t^2)^2}dt\\ &=\int\frac{1}{(1+t)^2 (1-t)^2}dt\\ &=\frac{1}{4}\int\left\lbrace\frac{1}{1+t}+\frac{1}{1-t}+\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right\rbrace dt\\ &=\frac{1}{4}\left\lbrace\log\left|\frac{1+t}{1-t}\right|+\frac{2t}{(1+t)(1-t)}\right\rbrace+C\\ &=\frac{1}{4}\left\lbrace\log\left|\frac{1+t}{1-t}\right|+\frac{2t}{(1+t)(1-t)}\right\rbrace+C\\ &=\frac{1}{4}\left(\log\frac{1+\sin{x}}{1-\sin{x}}+\frac{2\sin{x}}{\cos^2{x}}\right)+C\qquad(答) \end{align*}

\(\displaystyle \fbox{73}***\qquad\int^{\frac{\pi}{2}}_0\frac{\cos x}{\sin x+2\cos x}dx\) \begin{align*} %Q73 J&=\int^{\frac{\pi}{2}}_0\frac{(\sin x+2\cos x)'}{\sin x+2\cos x}dx\\ &=\int^{\frac{\pi}{2}}_0\frac{\cos x-2\sin x}{\sin x+2\cos x}dx\\ &=\left[\log|\sin x+2\cos x\right]^{\frac{\pi}{2}}_0\\ &=-\log 2\\ 5I&=5×\int^{\frac{\pi}{2}}_0\frac{\cos x}{\sin x+2\cos x}dx\\ &=\int^{\frac{\pi}{2}}_0 \frac{\cos x-2\sin x}{\sin x+2\cos x}+2dx\\ &=\left[-\log 2+2x\right]^{\frac{\pi}{2}}_0\\ &=\pi-\log 2\\ \therefore I&=\frac{\pi}{5}-\frac{1}{5}\log 2\qquad(答)\\ \end{align*} 三角関&数の合成を用いた別解を示す. \begin{align*} \cos\alpha&=\frac{1}{\sqrt{5}}\quad\\ \sin\alpha&=\frac{2}{\sqrt{5}} \end{align*} \begin{align*} x&=t-\alpha\qquad\quad\\ dx&=dt \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0&\rightarrow & \frac{\pi}{2} \\ \hline t & \alpha&\rightarrow & \frac{\pi}{2}+\alpha \\ \hline \end{array} と置換して, \begin{align*} \int^{\frac{\pi}{2}}_0\frac{\cos x}{\sqrt{5}(\frac{1}{\sqrt{5}}\sin x+\frac{2}{\sqrt{5}}\cos x)}dx&=\int^{\frac{\pi}{2}}_0\frac{\cos x}{\sqrt{5}(\cos\alpha\sin x+\sin\alpha\cos x)}dx\\ &=\int^{\frac{\pi}{2}}_0\frac{\cos x}{\sqrt{5}(\sin(x+\alpha))}dx\\ &=\int^{\frac{\pi}{2}+\alpha}_\alpha\frac{\cos(t-\alpha)}{\sqrt{5}\sin t}dt\\ &=\int^{\frac{\pi}{2}+\alpha}_\alpha\frac{\cos t\cos\alpha+\sin t\sin\alpha}{\sqrt{5}\sin t}dt\\ &=\int^{\frac{\pi}{2}+\alpha}_\alpha\frac{\frac{1}{\sqrt{5}}\cos t+\frac{2}{\sqrt{5}}\sin t}{\sqrt{5}\sin t}dt\\ &=\frac{1}{5}\int^{\frac{\pi}{2}+\alpha}_\alpha\frac{\cos t+2\sin t}{\sin t}dt\\ &=\frac{1}{5}\left\lbrace\int^{\frac{\pi}{2}+\alpha}_\alpha\frac{\cos t}{\sin t}dt+2\int^{\frac{\pi}{2}+\alpha}_\alpha dt\right\rbrace\\ &=\frac{1}{5}\left\lbrace\log{\sin(\alpha+\frac{\pi}{2})}-\log{\sin\alpha}+2(\frac{\pi}{2})\right\rbrace\\ &=\frac{1}{5}\left\lbrace\log(\frac{1}{\sqrt{5}})-\log(\frac{2}{\sqrt{5}})+\pi\right\rbrace\\ &=\frac{\pi}{5}-\frac{1}{5}\log2\qquad(答) \end{align*}