積分ガチャの解答

$\displaystyle\fbox{1}\qquad I=\int \frac{x^3}{(x+1)(x+2)} dx$ $$ \begin{align*} x^3&=(x^2+3x+2)(x-3)+7x+6\\ \end{align*} $$より $$ \begin{align*} \frac{x^3}{(x+1)(x+2)}&=x-3+\frac{7x+6}{(x+1)(x+2)}\\ \end{align*} $$ から $$ \begin{align*} \therefore I&=\int\left(x-3+\frac{-1}{x+1}+\frac{8}{x+2}\right)dx\\ &=\frac{x^2}{2}-3x-\log|x+1|+8\log|x+2| + C\qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{2}\qquad\int_{0}^{2\pi} \sqrt{1+\cos x} dx$ $$ \begin{align*} %問2 \sqrt{1+\cos x}&=\sqrt{2\cos^2\frac{x}{2}}\\ &=\sqrt{2}|\cos\frac{x}{2}|\\ &=\begin{cases} \sqrt{2}\cos\frac{x}{2} & (0\leq x \leq \pi)\\ -\sqrt{2}\cos\frac{x}{2} & (\pi \leq x \leq 2\pi) \end{cases} \end{align*} $$ $$ \begin{align*} \int_{0}^{2\pi} \sqrt{1+\cos x} dx &=\int_{0}^{\pi} \sqrt{1+\cos x} dx +\int_{\pi}^{2\pi} -\sqrt{1+\cos x} dx\\ &=\sqrt{2}\left\lbrace \left[2\sin\frac{x}{2}\right]^{\pi}_0 -\left[2\sin\frac{x}{2}\right]^{2\pi}_{\pi}\right\rbrace\\ &=\sqrt{2}{(2-0)-(0-2)}\\ &=4\sqrt{2}\qquad(答) \end{align*} $$

$\displaystyle\fbox{3}\qquad\int^1_0 \sqrt{\frac{1-x}{1+x}}dx$ $$ \begin{flalign*} t^2&=\frac{1-x}{1+x}\\ x(1+x^2)&=1-t^2\\ x&=\frac{1-t^2}{1+t^2}\\ \frac{dx}{dt}&=-\frac{4t^2}{1+t^2}\\ \\ \end{flalign*} $$ 以下のように置換すると$$ \begin{align*} t&=\sqrt{\frac{1-x}{1+x}} \qquad \end{align*} $$$$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline t & 1 & \rightarrow & 0 \\ \hline \end{array} $$$$ \begin{align*} \frac{dx}{dt}&=-\frac{4t^2}{1+t^2}\qquad\\ \end{align*} $$ とかけ, $$ \begin{align*} \int^1_0 \sqrt{\frac{1-x}{1+x}}dx&=\int^0_1t\left\lbrace-\frac{4t}{(1+t^2)^2}dt\right\rbrace\\ &=4\int^1_0\frac{t^2}{(1+t^2)^2}dt \end{align*} $$ 以下のように置換すると $$ \begin{align*} t&=\tan\theta \end{align*} $$$$ \begin{array}{|c||ccc|} %増減表 \hline t & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} $$$$ \begin{align*} dt=\frac{1}{\cos^2\theta}d\theta \qquad \\ \end{align*} $$ とかけ, $$ \begin{align*} &=4\int_{0}^{\frac{\pi}{4}}\tan^2\theta\cdot \cos^4\theta\cdot \frac{1}{\cos^2\theta}d\theta\\ &=\int_{0}^{\frac{\pi}{4}}\sin^2\theta d\theta\\ &=\int_{0}^{\frac{\pi}{4}}\frac{1-\cos2\theta}{2}d\theta\\ &=\frac{1}{2}\left[\theta-\frac{1}{2}\sin2\theta\right]^{\frac{\pi}{4}}_0\\ &=\frac{{\pi}}{8}-\frac{1}{4} \qquad(答)\\ \\ \end{align*} $$

$\displaystyle\fbox{4}\qquad\int x\sqrt{x+1} dx$ $$ \begin{flalign*} x\sqrt{x+1}&=(x+1)\sqrt{x+1}-\sqrt{x+1}\\ &=(x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}\\ \int x\sqrt{x+1} dx&=\int (x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}dx\\ &=\frac{2}{5}(x+1)^{\frac{5}{2}}-\frac{2}{3}(x+1)^{\frac{3}{2}} \qquad(答) \end{flalign*} $$

$\displaystyle\fbox{5}\qquad\int \frac{1}{x^2}(1+\frac{2}{x})^2 dx$ $$ \begin{align*} %Q5 \left\lbrace1+\frac{2}{x}\right\rbrace'&=-\frac{2}{x^2}より\\ \int -\frac{1}{2}(1+\frac{2}{x})'(1+\frac{2}{x})^2 dx&=-\frac{1}{6}(1+\frac{2}{x})^3+C \qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{6}\qquad\int^1_0 \sqrt{1+2\sqrt{x}} dx$ $$ \begin{flalign*} t&=\sqrt{1+2\sqrt{x}}\\ \end{flalign*} $$ と置換し, $$ \begin{flalign*} x&=(\frac{t^2-1}{2})^2\\ &=\frac{1}{4}(t^2-1)^2\\ \end{flalign*} $$ となり $$ \begin{align*} x&=\frac{1}{4}(t^2-1)^2\qquad \end{align*} $$ と置換すると, $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline t & 1 & \rightarrow & \sqrt{3} \\ \hline \end{array} $$ $$ \begin{align*} dx=t(t^2-1)dt \qquad\\ \end{align*} $$ とかけ, $$ \begin{align*} \int^1_0 \sqrt{1+2\sqrt{x}} dx&=\int^{\sqrt{3}}_1t\cdot t(t^2-1)dt\\ &=\left[\frac{1}{5}t^5-\frac{1}{3}t^3\right]^{\sqrt{3}}_1\\ &=\frac{4}{5}\sqrt{3}+\frac{2}{15}\qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{7}* \qquad\int_{1}^{2}\frac{1}{2^x}dx$ $$\\$$ 対数微分法より $$ \begin{align*} y&=2^{-x}\\ \log y&=-x\log 2\\ \frac{y'}{y}&=-\log 2\qquad\\ y'&=-2^{-x}\log 2\\ \end{align*} $$ $$ \begin{align*} (2^{-x})'=-2^{-x}\log 2\\ 2^{-x}=(-\frac{2^{-x}}{\log 2})'\\ \end{align*} $$ とかけるから, $$ \begin{align*} \int_{1}^{2}\frac{1}{2^x}dx=\left[-\frac{2^{-x}}{\log2}\right]^2_1&=-\frac{1}{\log2}(2^{-2}-2^{-1})\\ &=\frac{1}{4\log2}\qquad(答) \end{align*} $$

$\displaystyle\fbox{8}*\qquad\int^1_0\frac{1}{x^2-2x+4}$ $$ \begin{align*} \int^1_0\frac{1}{x^2-2x+4}=\int^1_0\frac{1}{(x-1)^2 +3}\\ \end{align*} $$ $$ \begin{align*} &x-1=\sqrt{3}\tan\theta\\ &dx=\sqrt{3}\frac{1}{\cos^2\theta} d\theta\; \end{align*} $$ と置換すると, $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & -\frac{\pi}{6} & \rightarrow & 0 \\ \hline \end{array} $$ $$ \begin{align*} &=\int^0_{-\frac{\pi}{6}}\frac{\cos^2\theta}{3}\cdot \sqrt{3}\cdot \frac{1}{\cos^2\theta}d\theta\\ &=\frac{\sqrt{3}}{3}\cdot\frac{\pi}{6}\\ &=\frac{\sqrt{3}}{18}\pi\qquad(答) \end{align*} $$

$\displaystyle\fbox{9}****\qquad\int \frac{1}{\sqrt{1+x^2}}dx$ $$ \begin{align*} x&=\frac{e^t-e^{-t}}{2}\\ t&=\log \left(x+\sqrt{x^2+1}\right)\\ dx&=\frac{e^t+e^{-t}}{2}dt\\ \end{align*} $$ とおくと, $$ \begin{align*} 2x=e^t-e^{-t}\\ (e^t)^2-2xe^t-1=0\\ e^t=x+\sqrt{x^2+1}\\ \end{align*} $$ $$ \begin{align*} \int \frac{1}{\sqrt{1+x^2}}dx&=\int\frac{1}{\sqrt{1+\frac{\left(e^t-e^{-t}\right)^2}{4}}}\frac{e^t+e^{-t}}{2}dt\\ &=\int\frac{1}{\frac{e^t+e^{-t}}{2}}\cdot\frac{e^t+e^{-t}}{2}dt\\ &=\int dt\\ &=t+C\\ &=\log\left(x+\sqrt{x^2+1}\right)+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{10}**\qquad\int x\sqrt{3x-5}dx$ $$ \begin{align*} t&=\sqrt{3x-5}\qquad\\ x&=\frac{t^2+5}{3}\qquad\\ dx&=\frac{2}{3}tdt\qquad\\ \end{align*} $$ と置換すると, $$ \begin{align*} \int \frac{t^2+5}{3}\cdot t\cdot\frac{2}{3}tdt&=\frac{2}{9}\int t^4+5t^2dt\\ &=\frac{2}{9}\left(\frac{1}{5}t^5+\frac{5}{3}t^3\right)+C\\ &=\frac{2}{45}(3x-5)\sqrt{3x-5}(3x+\frac{10}{3})+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{11}*\qquad\int^2_0 \frac{x}{4+x^2}dx$ $$ \begin{align*} %Q11 u=4+x^2,\quad du&=2xdxと置換\\ I&=\int^5_4\frac{\frac{1}{2}du}{u}\\ &=\left[\frac{1}{2}\log|u|\right]^5_4\\ &=\frac{1}{2}(\log 5-\log 4)\qquad(答) \end{align*} $$

$\displaystyle\fbox{12}***\qquad\int\frac{x}{(x^2+2)(x^2+3)}dx$ $$ \begin{align*} %Q12 \int\frac{x}{(x^2+2)(x^2+3)}dx&=\int x\left(\frac{1}{x^2+2}-\frac{1}{x^2+3}\right)dx\\ &=\int\left(\frac{x}{x^2+2}-\frac{x}{x^2+3}\right)dx\\ &=\int\left\lbrace\frac{1}{2}\cdot\frac{(x^2+2)'}{x^2+2}-\frac{1}{2}\frac{(x^2+3)'}{x^2+3}\right\rbrace dx\\ &=\frac{1}{2}\log\frac{x^2+2}{x^2+3}+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{13}*\qquad\int^1_{-1}\frac{x^3}{1+x^2}dx$ $$ \begin{align*} %Q13 \frac{x^3}{1+x^2}は奇関数である為,\\ \int^1_{-1}\frac{x^3}{1+x^2}dx=0\qquad(答) \end{align*} $$

$\displaystyle\fbox{14}***\qquad\int\frac{4(3+3x-x^2)}{(x-1)^2(x+1)}dx$ $$ \begin{align*} %Q14 \int\frac{4(3+3x-x^2)}{(x-1)^2(x+1)}dx&=\int\left\lbrace-\frac{3}{x-1}+\frac{10}{(x-1)^2}-\frac{1}{x+1}\right\rbrace dx\\ &=\log|\frac{1}{(x-1)^3(x+1)}-\frac{10}{x-1}+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{15}****\quad\int^1_0\sqrt{\frac{x}{1+x}}dx$ $$ \begin{align*} x&=\tan^2\theta\quad\\ dx&=\frac{2\tan x}{\cos^2\theta}d\theta\;\; \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} $$ と置換すると, $$ \begin{align*} \int^1_0\sqrt{\frac{x}{1+x}}dx&=\int^{\frac{\pi}{4}}_0\sqrt{\frac{\tan^2\theta}{1+\tan^2\theta}}\cdot 2\tan\theta\frac{1}{\cos^2\theta}d\theta\\ &=\int^{\frac{\pi}{4}}_0 2\tan\theta\cdot\cos\theta\cdot\frac{\sin\theta}{\cos^3\theta}d\theta\\ &=\int^{\frac{\pi}{4}}_0\frac{2\sin^2\theta}{\cos^3\theta}d\theta\\ &=\int^{\frac{\pi}{4}}_0\sin\theta\cdot\left(\frac{1}{\cos^2\theta}\right)'d\theta\\ &=\left[\sin\theta\cdot\frac{1}{\cos^2\theta}\right]^{\frac{\pi}{4}_0}-\int^{\frac{\pi}{4}}_0\frac{1}{\cos^2\theta}d\theta\\ &=\sqrt{2}-\int^{\frac{\pi}{4}}_0 \frac{\cos\theta}{\cos^2\theta}d\theta\\ &=\sqrt{2}-\int^{\frac{\pi}{4}}_0 \frac{\cos\theta}{1-\sin^2\theta}d\theta\\ \end{align*} $$ ここで $$ \begin{align*} t&=\sin\theta\quad\\ dt&=\cos\theta d\theta\qquad \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} $$ と置換すると, $$ \begin{align*} &=\sqrt{2}-\int^{\frac{1}{\sqrt{2}}}_0 \frac{1}{1-t^2}dt\\ &=\sqrt{2}-\int^{\frac{1}{\sqrt{2}}}_0 \left(\frac{1}{1+t}+\frac{1}{1-t}\right)dt\\ &=\sqrt{2}-\frac{1}{2}\left[\log|1+t|-\log|1-t|\right]^{\frac{1}{\sqrt{2}}}_0\\ &=\sqrt{2}-\frac{1}{2}\log\frac{\sqrt{2}+1}{\sqrt{2}-1}\\ &=\sqrt{2}-\frac{1}{2}\log(\sqrt{2}+1)^2\\ &=\sqrt{2}-\log(\sqrt{2}+1)\qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{16}***\quad\int\sqrt{x\sqrt{x\sqrt{x\cdots}}}dx$ $$ \begin{align*} %Q16 \sqrt{x\sqrt{x\sqrt{x\cdots}}}&=(x(x(x\cdots)^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}}\\ &=x^{\frac{1}{2}}x^{\frac{1}{4}}x^{\frac{1}{8}}\cdots\\ &=x^{\sum\limits^{\infty}_{n=1}\frac{1}{2^n}}=x\\ \therefore\int\sqrt{x\sqrt{x\sqrt{x\cdots}}}dx&=\int xdx =\frac{1}{2}x^2+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{17}***\qquad\int^1_{\frac{1}{2}}x\sqrt{\frac{1}{x}-1}dx$ $$ \begin{align*} x-\frac{1}{2}&=\frac{1}{2}\sin\theta\\ dx&=\frac{1}{2}\cos\theta d\theta \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & \frac{1}{2} & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{2} \\ \hline \end{array} $$ と置換すると, $$ \begin{align*} \int^1_{\frac{1}{2}}&x\sqrt{\frac{1}{x}-1}dx\\ &=\int^1_{\frac{1}{2}}\sqrt{x-x^2}dx\\ &=\int^1_{\frac{1}{2}}\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}dx\\ &=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^2\theta}\cdot\frac{1}{2}\cos\theta d\theta\\ &=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\cos^2\theta d\theta\\ &=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{1+\cos 2\theta}{2}d\theta\\ &=\frac{1}{4}\left[\frac{\theta}{2}+\frac{\sin 2\theta}{4}\right]^{\frac{\pi}{2}}_0\\ &=\frac{\pi}{16}\qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{18}**\qquad\int\frac{1}{x(x+1)(x+2)}dx$ $$ \begin{align*} %Q18 \int\frac{1}{x(x+1)(x+2)}dx&=\int\left(\frac{1}{2x}-\frac{1}{x+1}+\frac{1}{2(x+2)}\right)dx\\ &=\frac{1}{2}\log|x|-\log|x+1|+\frac{1}{2}\log|x+2|+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{19}**\qquad\int\frac{x}{1-x^2}dx$ $$ \begin{align*} %Q19 \int\frac{x}{1-x^2}dx&=-\frac{1}{2}\int\frac{(1-x^2)'}{1-x^2}dx\\ &=-\frac{1}{2}\log|1-x^2|+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{20}***\qquad\int^2_{\sqrt{2}}\frac{1}{x\sqrt{x^2-1}}dx$ $$ \begin{align*} t&=\sqrt{x^2-1}\\ dt&=\frac{x}{\sqrt{x^2-1}}dx\;\; \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & \sqrt{2} & \rightarrow &2 \\ \hline t & 1 & \rightarrow & \sqrt{3} \\ \hline \end{array} $$ $$ \begin{align*} t&=\tan\theta\\ dt&=\frac{1}{\cos^2\theta}d\theta\quad\;\;\; \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline t & 1 & \rightarrow & \sqrt{3} \\ \hline \theta & \frac{\pi}{4} & \rightarrow & \frac{\pi}{3} \\ \hline \end{array} $$ と置換して $$ \begin{align*} \int^2_{\sqrt{2}}\frac{1}{x\sqrt{x^2-1}}dx&=\int^{\sqrt{3}}_1\frac{1}{t^2+1}dt\\ &=\int^{\frac{\pi}{3}}_{\frac{\pi}{4}}d\theta\\ &=\frac{\pi}{12}\qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{21}**\qquad\int\frac{1}{x\sqrt{1-x^2}}dx$ $$ \begin{align*} t&=\sqrt{1-x^2}\\ dt&=-\frac{x}{\sqrt{1-x^2}}dx\\ \end{align*} $$ と置換すると, $$ \begin{align*} \int\frac{1}{x\sqrt{1-x^2}}dx&=\int\frac{1}{t^2-1}dt\\ &=\frac{1}{2}\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt\\ &=\frac{1}{2}\log|t-1|-\frac{1}{2}\log|t+1|+C\\ &=\frac{1}{2}\log|\sqrt{1-x^2}-1|\\ &-\frac{1}{2}\log|\sqrt{1-x^2}+1|+C\qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{22}*\qquad\int_{0}^{\pi}\left|\cos x-\cos2x\right|dx$ $$ \begin{align*} %Q22 f(x)&=\cos x-\cos2xとおいて符号を調べる\\ &=\cos x-(2\cos^2x-1)\\ &=-2\cos^2x+\cos x+1\\ &=(1-\cos x)(2\cos x+1)\\ 積分区&間である 0\leq x \leq\pi の時, 1-\cos x\geq0が成立するので\\ 2\cos x&+1=0となる x=\frac{2}{3}\pi の前後で正から負に変わる \end{align*} $$ $$ \begin{align*} \int_{0}^{\pi}\left|f(x)\right|dx&=\int_{0}^{\frac{2}{3}\pi}f(x)dx+\int_{\frac{2}{3}\pi}^{\pi}-f(x)dx\\ &=\int_{0}^{\frac{2}{3}\pi}f(x)dx+\int_{\pi}^{\frac{2}{3}\pi}f(x)dx\\ &=\left[\sin x-\frac{1}{2}\sin2x\right]^{\frac{2}{3}\pi}_0+\left[\sin x-\frac{1}{2}\sin2x\right]^{\frac{2}{3}\pi}_{\pi}\\ &=2\left\lbrace\frac{\sqrt{3}}{2}-\frac{1}{2}\left(-\frac{\sqrt{3}}{2}\right)\right\rbrace\\ &=\frac{3}{2}\sqrt{3}\qquad(答) \end{align*} $$

指数対数関数の積分

$\displaystyle\fbox{23}**\qquad\int_{1}^{e}\sqrt{x}\log xdx$ $$ \begin{align*} %23 \int_{1}^{e}x^{\frac{1}{2}}\log xdx&=\left[\frac{2}{3}x^{\frac{3}{2}}\log x\right]-\int_{1}^{e}\frac{2}{3}x^{\frac{3}{2}}\cdot x^{-1}dx\\ \qquad&=\frac{2}{3}e^{\frac{3}{2}}-\frac{2}{3}\int_{1}^{e}x^{\frac{1}{2}}dx\\ \qquad&=\frac{2}{9}e\sqrt{e}+\frac{4}{9}\qquad(答) \end{align*} $$

$\displaystyle\fbox{24}*\qquad\int_{0}^{2}\frac{3x^3+12x+1}{x^2+4}dx$ $$\\$$ 分母と同じ形を作って, 和の形をつくる. $$ \begin{align*} x&=2\tan\theta\\ dx&=2\frac{1}{\cos^2\theta}d\theta\quad \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 2 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} $$ と置換すると, \begin{align*} \int_{0}^{2}\frac{3x^2+12x+1}{x^2+4}dx&=\int_{0}^{2}\frac{3x(x^2+4)+1}{x^2+4}dx\\ &=3\int_{0}^{2}xdx+\int_{0}^{2}\frac{1}{x^2+4}dx\\ &=3\left[\frac{x^2}{2}\right]^2_0+\int_{0}^{\frac{\pi}{4}}\frac{1}{4(\tan^2\theta+1)}\frac{2d\theta}{\cos^2\theta}\\ &=6+\frac{1}{2}\int_{0}^{\frac{\pi}{4}}d\theta\\ &=6+\frac{\pi}{8}\qquad(答) \end{align*}

$\displaystyle\fbox{25}**\qquad\int\left(\log x\right)^2dx$ $$ \begin{align*} %Q25 \int\left(\log x\right)^2dx&=\int 1\cdot\left(\log x\right)^2dx\\ &=x\left(\log x\right)^2-\int x\cdot 2\log x\cdot\frac{1}{x}dx\\ &=x\left(\log x\right)^2-2\int 1\cdot \log x dx\\ &=x\left(\log x\right)^2-2\left(x\log x-\int x \cdot\frac{1}{x} dx\right)\\ &=x\left(\log x\right)^2-2\left(x\log x-x\right)+C\quad(答)\\ \end{align*} $$

$\displaystyle\fbox{26}\qquad\int\frac{1}{x\left(1+\log x\right)}dx$ $$ \begin{align*} t&=\log x\qquad\\ dt&=\frac{1}{x}\qquad\\ dx&=xdt\qquad \end{align*} $$ と置換すると, \begin{align*} \int\frac{1}{x\left(1+\log x\right)}dx&=\int\frac{1}{1+t}dt\\ &=\log |1+t|+C\\ &=\log |1+\log x|+C\qquad(答) \end{align*}

$\displaystyle\fbox{27}**\qquad\int\frac{x}{\cos^2x}dx$ $$ \begin{align*} %Q27 \int\frac{x}{\cos^2x}dx&=\int x\left(\tan x\right)'dx\\ &=x\tan x-\int\tan xdx\\ &=x\tan x+\int\frac{(\cos x)'}{\cos x}dx\\ &=x\tan x+\log|\cos x|+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{28}*\qquad\int\log_2xdx$ $$ \begin{align*} %Q28 \int\log_2xdx&=\int\frac{\log x}{\log 2}\\ &=\frac{1}{\log 2}\int\log xdx\\ &=\frac{1}{\log 2}\left(x\log x-\int x\cdot\frac{1}{x}dx\right)\\ &=\frac{1}{\log 2}\left(x\log x-x\right)+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{29}***\qquad\int \tan x \log\left( \cos^2x\right) dx $ $$ \begin{align*} t&=\cos x\\ dt&=-\sin xdx\\ \end{align*} $$ と置換すると, \begin{align*} \int \tan x \log\left( \cos^2x\right) dx&=\int \frac{\sin x}{\cos x}\log |\cos x|^2dx \\ &=-\int \frac{1}{t}\log |t|^2dt\\ &=-2\int \left(\log|t|\right)'\log|t|dt\\ &=-2\cdot\frac{1}{2}\left(\log|t|\right)^2+C\\ &=-\left(\log|\cos x|\right)^2+C\qquad(答) \end{align*}

$\displaystyle\fbox{30}***\qquad\int 2^{\log x} dx $ $$\\$$ [導出] $$ \begin{align*} a^{\log_b x}&=\left(b^{\log_b a}\right)^{\log_b x}\\ &=b^{\log_b a\log_b x}\\ &=\left(b^{\log_b x}\right)^{\log_b a}\\ &=x^{\log_b a} \end{align*} $$ \begin{align*} \int 2^{\log x} dx&=\int x^{\log2}dx\\ &=\frac{1}{\log2 +1}x^{\log2 +1}+C\\ &=\frac{x\cdot 2^{\log x}}{\log 2 +1}+C\qquad(答) \end{align*}

$\displaystyle\fbox{31}**\qquad\int \frac{\log\left(\log x\right)}{x\log x} dx$ $$ \begin{align*} t&=\log x\\ dt&=\frac{1}{x}dx\\ \end{align*} $$ と置換すると, \begin{align*} &\int \frac{\log\left(\log x\right)}{x\log x} dx \\ &=\int\frac{\log t}{t}dt\\ &=\int\left(\log t\right)\cdot\left(\log t\right)'dt\\ &=\frac{1}{2}\left(\log x\right)^2 +C\\ &=\frac{1}{2}\lbrace\log(\log x)\rbrace^2 +C\qquad(答) \end{align*}

$\displaystyle\fbox{32}**\qquad\int\frac{1}{x\left(\log x\right)^2}dx$ $$ \begin{align*} %Q32 \int\frac{1}{x\left(\log x\right)^2}dx&=\int \left(\log x\right)'\left(\log x\right)^{-2}dx\\ &=-\left(\log x\right)^{-1}+C\\ &=-\frac{1}{\log x}+C\qquad(答)\\ \end{align*} $$

$\displaystyle\fbox{33}*\qquad\int \sqrt{e^x}dx$ $$ \begin{align*} %Q33 \int \sqrt{e^x}dx&=\int \left(e^x\right)^{\frac{1}{2}}dx\\ &=\int e^{\frac{x}{2}}dx\\ &=2e^{\frac{x}{2}}+C\qquad(答) \end{align*} $$

$\displaystyle\fbox{34}*****\quad\int^1_{-1}\frac{x^2}{1+e^x}dx$ $$ \begin{align*} %Q34 x=-t,\quad dx&=-dtと置換する.\\ \int^1_{-1}\frac{x^2}{1+e^x}dx&=\int^0_{-1}\frac{x^2}{1+e^x}dx+\int^1_0\frac{x^2}{1+e^x}dx\\ &=\int^1_0\frac{t^2}{1+e^{-t}}dt+\int^1_0\frac{x^2}{1+e^x}dx\\ &=\int^1_0\frac{t^2 e^t}{1+e^{t}}dt+\int^1_0\frac{x^2}{1+e^x}dx\\ &=\int^1_0\frac{x^2\left(1+e^x\right)}{1+e^x}dx\\ &=\int^1_0 x^2dx\\ &=\frac{1}{3}\qquad(答) \end{align*} $$

$\displaystyle\fbox{35}**\qquad\int^{e^2}_e x^{\frac{1}{\log x}} dx$ $$ \begin{align*} %Q35 y&=x^{\frac{1}{\log x}}とおくと, \\ \log y&=\log x^{\frac{1}{\log x}}\\ &=\frac{1}{\log x}\log x=1\\ \therefore y&=e\\ \int^{e^2}_e x^{\frac{1}{\log x}}dx&=e\int^{e^2}_e dx\\ &=e\left(e^2-e\right)\\ &=e^3-e^2\qquad (答) \end{align*} $$

$\displaystyle\fbox{36}***\qquad\int\frac{1}{1-e^{-x}}$ $$ \begin{align*} %Q36 \int\frac{1}{1-e^{-x}}&=\int\frac{e^x}{e^x-1}\\ &=\int\frac{\left(e^x-1\right)'}{e^x-1}dx\\ &=\log|e^x-1|+C\qquad(答) \end{align*} $$

$\displaystyle \fbox{37}****\qquad\int x^x\left(\log x+1\right)dx$ $$ \begin{align*} %Q37 y=x^xとおくと&,\\ \log y&=\log x^x\\ &=x\log x\\ \frac{y'}{y}&=\log x+1\\ y'&=x^x\left(\log x+1\right)\\ \int x^x\left(\log x+1\right)dx&=\int \left(x^x\right)'dx\\ &=x^x+C\qquad(答) \end{align*} $$

$\displaystyle \fbox{38}***\qquad\int^{\frac{1}{2}\log 3}_0\frac{e^x}{1+e^{2x}}dx$ $$ \begin{align*} t&=e^x\qquad\qquad\\ dt&=e^x dx\qquad \end{align*} $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 &\rightarrow & \log{\sqrt{3}} \\ \hline t & 1 &\rightarrow &\sqrt{3} \\ \hline \end{array} \begin{align*} t&=\tan{\theta}\qquad\\ dt&=\frac{1}{\cos^2\theta}d\theta\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 1&\rightarrow &\sqrt{3} \\ \hline t & \frac{\pi}{4} &\rightarrow &\frac{\pi}{3} \\ \hline \end{array} と置換すると, \begin{align*} \int^{\frac{1}{2}\log 3}_0\frac{e^x}{1+e^{2x}}&=\int^{\frac{1}{2}\log 3}_0\frac{1}{1+\left(e^x\right)^2}e^x dx\\ &=\int^{\sqrt{3}}_1\frac{1}{1+t^2}dt\\ &=\int^{\frac{\pi}{3}}_{\frac{\pi}{4}}d\theta\\ &=\frac{\pi}{12}\qquad(答) \end{align*}

$\displaystyle \fbox{39}*\qquad\int e^{e^x+x}dx$ $$ \begin{align*} %Q39 \int e^{e^x+x}dx&=\int e^{e^x}\cdot e^x dx\\ &=\int \left(e^{e^x}\right)'dx\\ &=e^{e^x}+C\qquad(答) \end{align*} $$

$\displaystyle \fbox{40}****\qquad\int^1_0 \log\left(x^2+1\right)dx$ \begin{align*} x&=\tan\theta\quad\\ dx&=\frac{1}{\cos^2\theta} d\theta\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 1 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} と置換すると, \begin{align*} \int^1_0 \log\left(x^2+1\right)dx&= \left[x\log\left(x^2+1\right)\right]^1_0-\int^1_0 x\cdot\frac{2x}{x^2+1}dx\\ &=\log 2-2\int^1_0\frac{x^2}{x^2+1}dx\\ &=\log 2-2\left\lbrace\int^1_0 1dx-\int^1_0\frac{1}{x^2+1}dx\right\rbrace\\ &=\log 2-2+2\int_{0}^{\frac{\pi}{4}}d\theta\\ &=\log 2 -2+\frac{\pi}{2}\qquad(答) \end{align*}

$\displaystyle \fbox{41}***\qquad\int\frac{\log x}{x^2}dx$ \begin{align*} %Q41 \int\frac{\log x}{x^2}dx&=-\frac{1}{x}\log x+\int\frac{1}{x}\cdot\frac{1}{x}dx\\ &=-\frac{1}{x}\log x-\frac{1}{x}+C\\ &=-\frac{\log x+1}{x}+C\qquad(答) \end{align*}

$\displaystyle\fbox{42}***\qquad\int\left(\frac{\log x}{x}\right)^2dx $ \begin{align*} %Q42 t=\log x,\quad dt&=\frac{1}{x}dxと置換すると,\\ \int\left(\frac{\log x}{x}\right)^2dx&=\int\frac{t^2}{e^t}dt\\ &=\int t^2 e^{-t}dt\\ &=-t^2e^{-t}+2\int te^{-t}dt\\ &=-t^2e^{-t}+2\left(-te^{-t}+\int e^{-t}dt\right)\\ &=-t^2e^{-t}-2te^{-t}-2e^{-t}+C\\ &=-\frac{\left(\log x\right)^2+2\log x +2}{x}+C\qquad(答) \end{align*}

$\displaystyle\fbox{43}***\qquad\int\frac{1}{x\left(4-\left(\log x\right)^2\right)}dx $ \begin{align*} %Q43 t=\log x,\quad dt&=\frac{1}{x}dxと置換すると,\\ \int\frac{1}{x\left(4-\left(\log x\right)^2\right)}dx&=\int \frac{1}{4-t^2}dt\\ &=\int \frac{dt}{\left(2-t\right)\left(2+t\right)}\\ &=\frac{1}{4}\int\left(\frac{1}{2-t}+\frac{1}{2+t}\right)dt\\ &=\frac{1}{4}\left(-\log|2-t|+\log |2+t|\right)+C\\ &=\frac{1}{4}\left(-\log|2-\log x|+\log|2+\log x|\right)+C\qquad(答) \end{align*}

$\displaystyle \fbox{44}**\qquad\int\frac{\left(\log x+3\right)^2}{x}dx$ \begin{align*} %Q44 \int\frac{\left(\log x+3\right)^2}{x}dx&=\int\left(\log x+3\right)'\left(\log x+3\right)^2dx\\ &=\frac{1}{3}\left(\log x+3\right)^3+C\qquad(答) \end{align*}

$\displaystyle \fbox{45}****\qquad\int^{\sqrt{3}}_1\frac{1}{x^2}\log\sqrt{1+x^2}dx$ \begin{align*} %Q45 \int^{\sqrt{3}}_1&\frac{1}{x^2}\log\sqrt{1+x^2}dx=\frac{1}{2}\int^{\sqrt{3}}_1\left(-\frac{1}{x}\right)'\log\left(1+x^2\right)dx\\ &=\frac{1}{2}\left[\frac{1}{x}\log\left(1+x^2\right)\right]^{\sqrt{3}}_1-\frac{1}{2}\int^{\sqrt{3}}_1\left(-\frac{1}{x}\right)\frac{2x}{1+x^2}dx\\ &=\frac{1}{2}\left(-\frac{1}{\sqrt{3}}\log 4+\log 2\right)+\int_{1}^{\sqrt{3}}\frac{1}{1+x^2}dx\\ \end{align*} ここで \begin{align*} x&=\tan\theta\quad\\ dx&=\frac{1}{\cos^2\theta} d\theta\quad\; \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 1 & \rightarrow & \sqrt{3} \\ \hline \theta & \frac{\pi}{4} & \rightarrow & \frac{\pi}{3} \\ \hline \end{array} と置換すると, \begin{align*} &=\left(\frac{1}{2}-\frac{1}{\sqrt{3}}\right)\log 2+\frac{\pi}{12}\qquad(答) \end{align*}

$\displaystyle \fbox{46}***\qquad\int^2_0\frac{e^x}{e^x+e^{2-x}}dx$ \begin{align*} %Q46 \int^2_0\frac{e^x}{e^x+e^{2-x}}dx&=\int^2_0\frac{e^{2x}}{e^{2x}+e^2}dx\\ &=\frac{1}{2}\int^2_0\frac{\left(e^{2x}+e^2\right)'}{e^{2x}+e^2}dx\\ &=\frac{1}{2}\left[\log \left(e^{2x}+e^2\right)\right]^2_0\\ &=\frac{1}{2}\left\lbrace\log\left(e^4+e^2\right)-\log\left(1+e^2\right)\right\rbrace\\ &=\frac{1}{2}\log\frac{e^2\left(e^2+1\right)}{1+e^2}\\ &=1\qquad(答)\\\\ \end{align*} \begin{align*} \qquad\qquad～King \; Property～\qquad\qquad\qquad\qquad\\ \qquad\qquad\qquad\int^b_a f(x)dx=\int^b_a f(a+b-x)\qquad\qquad\qquad\\ \qquad\qquad\qquad2I=\int^b_a f(x)+f(a+b-x)dx\qquad\qquad\qquad\\ \end{align*} $King \; Property$を用いた別解を示す \begin{align*} I&=\int^2_0\frac{e^x}{e^x+e^{2-x}}dx\\ 2I&=\int^2_0\frac{e^x}{e^x+e^{2-x}}+\frac{e^{2-x}}{e^{2-x}+e^x}\\ &=\int^2_0 1dx=2\\ \qquad\qquad\qquad\qquad\therefore I&=1\qquad(答) \end{align*}

$\displaystyle \fbox{47}**\qquad\int\frac{3^x}{3^x+\log 3}dx$ \begin{align*} %Q47 \int\frac{3^x}{3^x+\log 3}dx&=\int\frac{1}{\log 3}\frac{\left(3^x+\log 3\right)'}{3^x+\log 3}dx\\ &=\frac{1}{\log 3}\log\left(3^x+\log 3\right)+C\qquad(答) \end{align*}

$\displaystyle \fbox{48}*****\int^4_2\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(x+3)}}dx$ \begin{align*} %Q48 I&=\int^4_2\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(x+3)}}dx\\ &=\int^4_2\frac{\sqrt{\log(x+3)}}{\sqrt{\log(x+3)}+\sqrt{\log(9-x)}}dx\\ 2I&=\int^4_2 1dx\\ &=2\\ \therefore I&=1 \end{align*}

$\displaystyle \fbox{49}*****\quad\int^1_0\frac{\log(1+x)}{1+x^2}dx$ \begin{align*} x&=\tan\theta\quad\\ dx&=\frac{1}{\cos^2\theta} d\theta\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} \begin{align*} \frac{\pi}{4}-\theta&=t\quad\\ d\theta&=-dt\qquad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline t & \frac{\pi}{4} & \rightarrow & 0 \\ \hline \end{array} と置換すると, \begin{align*} &=\int^{\frac{\pi}{4}}_0\log(1+\tan\theta)d\theta\\ &=\int^{\frac{\pi}{4}}_0\log(\frac{\cos\theta+\sin\theta}{\cos\theta})d\theta\\ &=\int^{\frac{\pi}{4}}_0\log(\frac{\sqrt{2}\cos(\frac{\pi}{4}-\theta)+\sin\theta}{\cos\theta})d\theta\\ &=\log\sqrt{2}\int^{\frac{\pi}{4}}_0 d\theta+\cancel{\int^{\frac{\pi}{4}}_0 \log\cos(\frac{\pi}{4}-\theta)d\theta}\cancel{-\int^{\frac{\pi}{4}}_0 \log\cos\theta d\theta}\\ &=\frac{\pi}{8}\log 2\qquad(答) \end{align*}

$\displaystyle \fbox{50}***\qquad\int\frac{\log x}{(x+1)^3}dx $ \begin{align*} %Q50 &=-\frac{1}{2}(x+1)^{-2}\log x+\frac{1}{2}\int\frac{1}{x(x+1)^2}dx\\ &=-\frac{\log x}{2(x+1)^2}+\frac{1}{2}\left\lbrace\log x-\log(x+1)+\frac{1}{x+1}\right\rbrace+C\\ &=\frac{1}{2}\left\lbrace\frac{x(x+2)}{(x+1)^2}\log x-\log(x+1)+\frac{1}{x+1}\right\rbrace +C\quad(答) \end{align*}

$\displaystyle\fbox{51}**\qquad\int^{e^e}_e\frac{\log x\cdot\log(\log x)}{x}dx$ \begin{align*} t&=\log x\quad\\ dt&=\frac{1}{x} dx \qquad\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & e & \rightarrow & e^e \\ \hline t & 1 & \rightarrow & e \\ \hline \end{array} と置換すると, \begin{align*} &=\int^e_1 t\cdot\log t dt\\ &=\left[\frac{1}{2}t^2\log t dt\right]^e_1 -\int^e_1\frac{1}{2}t^2\cdot\frac{1}{t}dt\\ &=\frac{1}{2}e^2-\frac{1}{2}\left[\frac{1}{2}t^2\right]^e_1\\ &=\frac{1}{2}e^2-\frac{1}{4}(e^2-1)\\ &=\frac{e^2-1}{4}\qquad(答)\\ \end{align*}

$\displaystyle \fbox{52}*\qquad\int^2_0 x\log(x+1)dx$ \begin{align*} %Q52 \int^2_0 x\log(x+1)dx=\left[\frac{x^2}{2}\log(x+1)-\int\frac{x^2}{2}\cdot\frac{1}{x+1}dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\int\frac{1}{2}(\frac{x^2-1+1}{x+1})dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\int\frac{1}{2}(\frac{(x+1)(x-1)+1}{x+1})dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\int\frac{1}{2}(x-1+\frac{1}{x+1})dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\frac{(x-1)^2}{4}-\frac{\log|x+1|}{2}\right]^2_0\\ &=\left(2\log 3-\frac{1}{4}-\frac{\log 3}{2}\right)-\left(0-\frac{1}{4}-0\right)\\ &=\frac{3\log 3}{2}\qquad(答) \end{align*}

$\displaystyle \fbox{53}**\qquad\int^1_0 x\log(x^2+1)dx$ \begin{align*} u&=x^2+1\qquad\\ du&=2xdx\qquad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x &0 & \rightarrow & 1 \\ \hline u & 1 & \rightarrow & 2 \\ \hline \end{array} と置換すると, \begin{align*} \int^1_0 x\log(x^2+1)dx&=\int^2_1\frac{1}{2}\log udu\\ &=\frac{1}{2}\left[u(\log u-1)\right]^2_1\\ &=\frac{1}{2}(2\log 2-2+1)\\ &=\frac{1}{2}(2\log 2-1)\qquad(答) \end{align*}

$\displaystyle \fbox{54}*\qquad\int^e_1\frac{\log x}{x^3}dx$ \begin{align*} %Q54 \int^e_1\frac{\log x}{x^3}dx&=\int^e_1\log x\cdot x^{-3}dx\\ &=\int^e_1\log x\cdot\left(\frac{x^{-2}}{-2}\right)'dx\\ &=\left[\log x\cdot\left(\frac{x^{-2}}{-2}\right)-\int\frac{1}{x}\cdot\left(\frac{x^{-2}}{-2}\right)dx\right]^e_1\\ &=\left[-\frac{1}{2x^2}\log x-\int\frac{x^{-3}}{-2}dx\right]^e_1\\ &=\left[-\frac{1}{2x^2}\log x-\frac{1}{4x^2}\right]^e_1\\ &=-\frac{1}{2e^2}-\frac{1}{4e^2}+\frac{1}{4}\\ &=-\frac{3}{4e^2}+\frac{1}{4}\qquad(答) \end{align*}

$\displaystyle \fbox{55}**\qquad\int\frac{1}{e^x(e^x+1)}dx$ \begin{align*} %Q55 \int\frac{1}{e^x(e^x+1)}dx&=\int\left(\frac{1}{e^x}-\frac{1}{e^x+1}\right)dx\\ &=-e^{-x}-\int\frac{1}{e^x+1}dx \end{align*} ここで \begin{align*} \quad t&=e^x\quad\\ \quad dt&=e^x dx\quad\\ \end{align*} と置換すると, \begin{align*} &=-e^{-x}-\int\frac{1}{t+1}\cdot \frac{dt}{t}\\ &=-e^{-x}-\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt\\ &=-e^{-x}-\log|t|+\log|t+1|+C\\ &=-e^{-x}-\log e^x+\log(e^x+1)+C\\ &=-e^{-x}-x+\log(e^x+1)+C\quad(答) \end{align*}

三角関数の積分

$\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*}