高等学校数学III 積分法/演習問題

問題

次の不定積分を計算せよ。

1. $\int {\left(\frac{x+2}{x}\right)}^{2}dx$
2. $\int (\sqrt{x}+1)(\sqrt{x}-2)dx$
3. $\int {(\sqrt{2x}+3)}^{2}dx$
4. $\int \frac{x-{\cos }^{2}x}{x{\cos }^{2}x}dx$
5. $\int ({\tan }^{2}x+\frac{1}{{\tan }^{2}x})dx$
6. $\int (e^x+\frac{1}{x})dx$
7. $\int 3^{x+2}dx$
8. $\int (x+2)(x^2+4x+1{)}^{3}dx$
9. $\int \frac{x+1}{(3x-1{)}^3}dx$
10. $\int {\cos }^{3}x\cdot {\sin }^{2}xdx$
11. $\int \frac{dx}{(1+\tan x){\cos }^{2}x}$
12. $\int (2x+1)e^{x^2+x+5}dx$
13. $\int \frac{e^{4x}}{e^{2x}+1}dx$
14. $\int x\sqrt{1-x}dx$
15. $\int \frac{3x-1}{\sqrt{x+1}}dx$
16. $\int \frac{x}{(x^2+4{)}^2}dx$
17. $\int \frac{x^2+1}{x^4-5x^2+4}dx$
18. $\int \frac{3x+2}{x(x+1{)}^3}dx$
19. $\int \frac{x}{\sqrt[3]{x+1}-1}dx$
20. $\int \frac{\cos x}{\sin x(\sin x+1)}dx$
21. $\int {(\tan x+\frac{1}{\tan x})}^{2}dx$
22. $\int {\tan }^{4}xdx$
23. $\int \frac{\sin 2x}{1+\sin x}dx$
24. $\int \frac{x}{1-\cos x}dx$
25. $\int \frac{1}{1-\sin x}dx$
26. $\int \frac{\sqrt{x}}{\sqrt[4]{x^3}+1}dx$
27. $\int \log (x^2-1)dx$
28. $\int \frac{e^{3x}}{e^x-1}dx$
29. $\int \frac{e^x}{e^x-e^{-x}}dx$
30. $\int \frac{dx}{3x^2-4x-2}$
31. $\int \frac{dx}{\sqrt{2x^2-4x+3}}$
32. $\int \frac{\tan x}{{\cos }^{3}x}dx$
33. $\int \frac{dx}{1+\cos x}$
34. $\int x{\tan }^{2}xdx$
35. $\int e^x\cos xdx$
36. $\int \sqrt{x^2-x-1}dx$
37. $\int \sin 2x\cdot \cos 3xdx$
38. $\int {(\sin x+\frac{1}{\sin x})}^{2}dx$
39. $\int \frac{1+{\cos }^{3}x}{{\cos }^{2}x}dx$
40. $\int \frac{dx}{1+\sin x}$
41. $\int \frac{dx}{e^x+1}$
42. $\int {(\log x)}^{2}dx$
43. $\int \frac{\sqrt{1-x^2}}{x}dx$
44. $\int \frac{dx}{x\sqrt{2x-x^2}}$

解答

以下の解答においては、問iにおいて計算すべき積分をI_iとおくことにする。

$\begin{array}{cc}{I}_1& =\int {\left(\frac{x+2}{x}\right)}^{2}dx\\ & =\int (1+\frac{4}{x}+\frac{4}{x^2})dx\\ & =x+4\log |x|-\frac{4}{x}+C\end{array}$

$\begin{array}{cc}{I}_2& =\int (\sqrt{x}+1)(\sqrt{x}-2)dx\\ & =\int (x-\sqrt{x}-2)dx\\ & =\frac{x^2}{2}-\frac{2}{3}x\sqrt{x}-2x+C\\ \end{array}$

$\begin{array}{cc}{I}_3& =\int {(\sqrt{2x}+3)}^{2}dx\\ & =\int (2x+6\sqrt{2x}+9)dx\\ & =x^2+4x\sqrt{2x}+9x+C\end{array}$

$\begin{array}{cc}{I}_4& =\int \frac{x-{\cos }^{2}x}{x{\cos }^{2}x}dx\\ & =\int (\frac{1}{{\cos }^{2}x}-\frac{1}{x})dx\\ & =\tan x-\log |x|+C\end{array}$

$\begin{array}{cc}{I}_5& =\int ({\tan }^{2}x+\frac{1}{{\tan }^{2}x})dx\\ & =\int ((\frac{1}{{\cos }^{2}x}-1)+(\frac{1}{{\sin }^{2}x}-1))dx\\ & =\tan x-\frac{1}{\tan x}-2x+C\end{array}$

$\begin{array}{cc}{I}_6& =\int (e^x+\frac{1}{x})dx\\ & =e^x+\log |x|+C\end{array}$

$\begin{array}{cc}{I}_7& =\int 3^{x+2}dx\\ & =\frac{3^{x+2}}{\log 3}+C\end{array}$

$\begin{array}{cc}{I}_8& =\int (x+2)(x^2+4x+1{)}^{3}dx\\ \end{array}$

$t=x^2+4x+1$とおくと$dt=(2x+4)dx=2(x+2)dx$なので、

$\begin{array}{cc}{I}_8& =\int \frac{t^3}{2}dt\\ & =\frac{t^4}{8}+C\\ & =\frac{1}{8}(x^2+4x+1{)}^4+C\end{array}$

$\begin{array}{cc}{I}_9& =\int \frac{x+1}{(3x-1{)}^3}dx\\ & =\int (\frac{1}{3(3x-1{)}^2}+\frac{4}{3(3x-1{)}^3})dx\\ & =-\frac{1}{9(3x-1)}-\frac{2}{9(3x-1{)}^2}+C\\ & =-\frac{3x+1}{9(3x-1{)}^2}+C\end{array}$

別解

$t=3x-1$と置く。 $x=\frac{t+1}{3},x+1=\frac{t+4}{3},\frac{dt}{dx}=3,dt=3dx$なので、

$\begin{array}{cc}{I}_9& =\frac{1}{9}\int \left(\frac{t+4}{t^3}\right)dt\\ & =\frac{1}{9}\int (\frac{1}{t^2}+\frac{4}{t^3})dt\\ & =-\frac{1}{9}(\frac{1}{t}+\frac{2}{t^2})+C\\ & =-\frac{1}{9}(\frac{1}{3x-1}+\frac{2}{(3x-1{)}^2})+C\\ & =-\frac{3x+1}{9(3x-1{)}^2}+C\end{array}$

$\begin{array}{cc}{I}_{10}& =\int {\cos }^{3}x\cdot {\sin }^{2}xdx\\ & =\int \cos x(1-{\sin }^{2}x){\sin }^{2}xdx\\ & =\int \cos x({\sin }^{2}x-{\sin }^{4}x)dx\\ \end{array}$

$t=\sin x$とおくと$dt=\cos xdx$なので、

$\begin{array}{cc}{I}_{10}& =\int (t^2-t^4)dt\\ & =\frac{t^3}{3}-\frac{t^5}{5}+C\\ & =\frac{1}{3}{\sin }^{3}x-\frac{1}{5}{\sin }^{5}x+C\end{array}$

$\begin{array}{cc}{I}_{11}& =\int \frac{dx}{(1+\tan x){\cos }^{2}x}\\ \end{array}$

$t=\tan x$とおくと$dt=\frac{dx}{{\cos }^{2}x}$なので、

$\begin{array}{cc}{I}_{11}& =\int \frac{dt}{1+t}\\ & =\log |1+t|+C\\ & =\log |1+\tan x|+C\end{array}$

$\begin{array}{cc}{I}_{12}& =\int (2x+1)e^{x^2+x+5}dx\\ \end{array}$

$t=x^2+x+5$とおくと$dt=(2x+1)dx$なので、

$\begin{array}{cc}{I}_{12}& =\int e^{t}dt\\ & =e^t+C\\ & =e^{x^2+x+5}+C\end{array}$

$\begin{array}{cc}{I}_{13}& =\int \frac{e^{4x}}{e^{2x}+1}dx\\ \end{array}$

$t=e^{2x}+1$とおくと$dt=2e^{2x}dx$なので、

$\begin{array}{cc}{I}_{13}& =\int \frac{t-1}{2t}dt\\ & =\frac{1}{2}\int (1-\frac{1}{t})dt\\ & =\frac{1}{2}(t-\log t)+C^{\prime }\\ & =\frac{1}{2}(e^{2x}+1-\log (e^{2x}+1))+C^{\prime }\\ & =\frac{1}{2}(e^{2x}-\log (e^{2x}+1))+C\end{array}$

$\begin{array}{cc}{I}_{14}& =\int x\sqrt{1-x}dx\\ \end{array}$

$t=\sqrt{1-x}$とおくと、$t^2=1-x,x=1-t^2,dx=-2tdt$なので、

$\begin{array}{cc}{I}_{14}& =-2\int (1-t^2)\cdot t\cdot tdt\\ & =-2\int (t^2-t^4)dt\\ & =-2(\frac{t^3}{3}-\frac{t^5}{5})+C\\ & =-\frac{2t^3}{15}(5-3t^2)+C\\ & =-\frac{2}{15}(1-x)\sqrt{1-x}(5-3(1-x))+C\\ & =-\frac{2}{15}(3x+2)(1-x)\sqrt{1-x}+C\\ \end{array}$

$\begin{array}{cc}{I}_{15}& =\int \frac{3x-1}{\sqrt{x+1}}dx\\ \end{array}$

$t=\sqrt{x+1}$とおくと、$t^2=x+1,x=t^2-1,dx=2tdt$なので、

$\begin{array}{cc}{I}_{15}& =\int \frac{3(t^2-1)-1}{t}\cdot 2tdt\\ & =2\int (3t^2-4)dt\\ & =2(t^3-4t)+C\\ & =2t(t^2-4)+C\\ & =2((x+1)-4)\sqrt{x+1}+C\\ & =2(x-3)\sqrt{x+1}+C\\ \end{array}$

$\begin{array}{cc}{I}_{16}& =\int \frac{x}{(x^2+4{)}^2}dx\\ \end{array}$

$t=x^2+4$とおくと$dt=2xdx$なので、

$\begin{array}{cc}{I}_{16}& =\int \frac{dt}{2t^2}\\ & =-\frac{1}{2t}+C\\ & =-\frac{1}{2(x^2+4)}+C\end{array}$

$\begin{array}{cc}{I}_{17}& =\int \frac{x^2+1}{x^4-5x^2+4}dx\\ & =\frac{1}{3}\int (\frac{5}{x^2-4}-\frac{2}{x^2-1})dx\\ & =\frac{1}{3}\int (\frac{5}{4}(\frac{1}{x-2}-\frac{1}{x+2})-(\frac{1}{x-1}-\frac{1}{x+1}))dx\\ & =\frac{5}{12}\log \left|\frac{x-2}{x+2}\right|-\frac{1}{3}\log \left|\frac{x-1}{x+1}\right|+C\\ & =\frac{1}{3}\log \left|\frac{x+1}{x-1}\right|-\frac{5}{12}\log \left|\frac{x+2}{x-2}\right|+C\\ \end{array}$

$\begin{array}{cc}{I}_{18}& =\int \frac{3x+2}{x(x+1{)}^3}dx\\ & =\int (\frac{1}{(x+1{)}^3}+\frac{2}{x(x+1{)}^2})dx\\ & =\int (\frac{1}{(x+1{)}^3}-\frac{2}{(x+1{)}^2}+\frac{2}{x(x+1)})dx\\ & =\int (\frac{1}{(x+1{)}^3}-\frac{2}{(x+1{)}^2}+\frac{2}{x}-\frac{2}{x+1})dx\\ & =-\frac{1}{2(x+1{)}^2}+\frac{2}{x+1}+2\log \left|\frac{x}{x+1}\right|+C\\ & =2\log \left|\frac{x}{x+1}\right|+\frac{2}{x+1}-\frac{1}{2(x+1{)}^2}+C\end{array}$

$\begin{array}{cc}{I}_{19}& =\int \frac{x}{\sqrt[3]{x+1}-1}dx\\ & =\int \frac{x({\left(\sqrt[3]{x+1}\right)}^2+\sqrt[3]{x+1}+1)}{x}dx\\ & =\int ((x+1{)}^{\frac{2}{3}}+(x+1{)}^{\frac{1}{3}}+1)dx\\ & =\frac{3}{5}(x+1{)}^{\frac{5}{3}}+\frac{3}{4}(x+1{)}^{\frac{4}{3}}+x+C\\ \end{array}$

$\begin{array}{cc}{I}_{20}& =\int \frac{\cos x}{\sin x(\sin x+1)}dx\\ \end{array}$

$t=\sin x$とおくと$dt=\cos xdx$なので、

$\begin{array}{cc}{I}_{20}& =\int \frac{dt}{t(t+1)}\\ & =\int (\frac{1}{t}-\frac{1}{t+1})dt\\ & =\log \left|\frac{t}{t+1}\right|+C\\ & =\log \left|\frac{\sin x}{\sin x+1}\right|+C\\ \end{array}$

$\begin{array}{cc}{I}_{21}& =\int {(\tan x+\frac{1}{\tan x})}^{2}dx\\ & =\int ({\tan }^{2}x+2+\frac{1}{{\tan }^{2}x})dx\\ & =\int ((\frac{1}{{\cos }^{2}x}-1)+2+(\frac{1}{{\sin }^{2}x}-1))dx\\ & =\tan x-\frac{1}{\tan x}+C\\ \end{array}$

$\begin{array}{cc}{I}_{22}& =\int {\tan }^{4}xdx\\ & =\int {\tan }^{2}x{\tan }^{2}xdx\\ & =\int {\tan }^{2}x(\frac{1}{{\cos }^{2}x}-1)dx\\ & =\int (\frac{{\tan }^{2}x}{{\cos }^{2}x}-{\tan }^{2}x)dx\\ & =\int (\frac{{\tan }^{2}x}{{\cos }^{2}x}-(\frac{1}{{\cos }^{2}x}-1))dx\\ & =\int \frac{{\tan }^{2}x}{{\cos }^{2}x}dx-(\tan x-x)\\ \end{array}$

$t=\tan x$とおくと$dt=\frac{1}{{\cos }^{2}x}dx$なので、

$\begin{array}{cc}{I}_{22}& =\int t^{2}dt-(\tan x-x)\\ & =\frac{1}{3}{t}^3-\tan x+x+C\\ & =\frac{1}{3}{\tan }^{3}x-\tan x+x+C\end{array}$

$\begin{array}{cc}{I}_{23}& =\int \frac{\sin 2x}{1+\sin x}dx\\ & =\int \frac{2\sin x\cos x}{1+\sin x}dx\end{array}$

$t=\sin x$とおくと$dt=\cos xdx$なので、

$\begin{array}{cc}{I}_{23}& =\int \frac{2t}{1+t}dt\\ & =2\int (1-\frac{1}{1+t})dt\\ & =2t-2\log |1+t|+C\\ & =2\sin x-2\log (1+\sin x)+C\end{array}$

$\begin{array}{cc}{I}_{24}& =\int \frac{x}{1-\cos x}dx\\ & =\int \frac{x}{2{\sin }^2\frac{x}{2}}dx\\ & =-\frac{x}{\tan \frac{x}{2}}+\int \frac{1}{\tan \frac{x}{2}}dx\\ & =-\frac{x}{\tan \frac{x}{2}}+\int \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}dx\end{array}$

$t=\sin \frac{x}{2}$とおくと$dt=\frac{1}{2}\cos \frac{x}{2}dx$なので、

$\begin{array}{cc}{I}_{24}& =-\frac{x}{\tan \frac{x}{2}}+2\int \frac{dt}{t}\\ & =-\frac{x}{\tan \frac{x}{2}}+2\log |t|+C\\ & =-\frac{x}{\tan \frac{x}{2}}+2\log |\sin \frac{x}{2}|+C\end{array}$

$\begin{array}{cc}{I}_{25}& =\int \frac{1}{1-\sin x}dx\\ & =\int \frac{1+\sin x}{1-{\sin }^{2}x}dx\\ & =\int (\frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x})dx\\ & =\tan x+\int \frac{\sin x}{{\cos }^{2}x}dx\end{array}$

$t=\cos x$とおくと$dt=-\sin xdx$なので、

$\begin{array}{cc}{I}_{25}& =\tan x-\int \frac{dt}{t^2}\\ & =\tan x+\frac{1}{t}+C\\ & =\tan x+\frac{1}{\cos x}+C\end{array}$

$\begin{array}{cc}{I}_{26}& =\int \frac{\sqrt{x}}{\sqrt[4]{x^3}+1}dx\end{array}$

$t=\sqrt[4]{x^3}$とおくと$dt=\frac{3}{4\sqrt[4]{x}}dx$なので、

$\begin{array}{cc}{I}_{26}& =\frac{4}{3}\int \frac{t}{t+1}dt\\ & =\frac{4}{3}\int (1-\frac{1}{t+1})dt\\ & =\frac{4}{3}t-\frac{4}{3}\log |t+1|+C\\ & =\frac{4}{3}\sqrt[4]{x^3}-\frac{4}{3}\log (\sqrt[4]{x^3}+1)+C\end{array}$

$\begin{array}{cc}{I}_{27}& =\int \log (x^2-1)dx\\ & =\int \log (x+1)(x-1)dx\\ & =\int (\log |x+1|+\log |x-1|)dx\\ & =(x+1)\log |x+1|-x+(x-1)\log |x-1|-x+C\\ & =(x+1)\log |x+1|+(x-1)\log |x-1|-2x+C\end{array}$

$\begin{array}{cc}{I}_{28}& =\int \frac{e^{3x}}{e^x-1}dx\end{array}$

$t=e^x$とおくと$dt=e^{x}dx$なので、

$\begin{array}{cc}{I}_{28}& =\int \frac{t^2}{t-1}dt\\ & =\int (t+1+\frac{1}{t-1})dt\\ & =\frac{1}{2}{t}^2+t+\log |t-1|+C\\ & =\frac{1}{2}{e}^{2x}+e^x+\log |e^x-1|+C\end{array}$

$\begin{array}{cc}{I}_{29}& =\int \frac{e^x}{e^x-e^{-x}}dx\\ \end{array}$

分子分母に$e^x$を掛けて、

$\begin{array}{cc}{I}_{29}& =\int \frac{e^{2x}}{e^{2x}-1}dx\end{array}$

$t=e^{2x}$とおくと$dt=2e^{2x}dx$なので、

$\begin{array}{cc}{I}_{29}& =\frac{1}{2}\int \frac{dt}{t-1}\\ & =\frac{1}{2}\log |t-1|+C\\ & =\frac{1}{2}\log |e^{2x}-1|+C\end{array}$

$\begin{array}{cc}{I}_{30}& =\int \frac{dx}{3x^2-4x-2}\\ & =\int \frac{3}{(3x-2-\sqrt{10})(3x-2+\sqrt{10})}dx\\ & =\int \frac{3}{2\sqrt{10}}(\frac{1}{3x-2-\sqrt{10}}-\frac{1}{3x-2+\sqrt{10}})dx\\ & =\frac{1}{2\sqrt{10}}(\log |3x-2-\sqrt{10}|-\log |3x-2+\sqrt{10}|)+C\\ & =\frac{1}{2\sqrt{10}}\log \left|\frac{3x-2-\sqrt{10}}{3x-2+\sqrt{10}}\right|+C\end{array}$

• $I_{30}$について補足

  2行目から3行目への変形では、$\frac{1}{3x-2-\sqrt{10}}-\frac{1}{3x-2+\sqrt{10}}=\frac{(3x-2+\sqrt{10})-(3x-2-\sqrt{10})}{(3x-2-\sqrt{10})(3x-2+\sqrt{10})}=\frac{2\sqrt{10}}{(3x-2-\sqrt{10})(3x-2+\sqrt{10})}$であることに注意せよ。

  3行目から4行目への変形では、$(\log |3x-2-\sqrt{10}|)=\frac{3}{3x-2-\sqrt{10}},(\log |3x-2+\sqrt{10}|)=\frac{3}{3x-2+\sqrt{10}}$であることに注意せよ。

$\begin{array}{cc}{I}_{31}& =\int \frac{dx}{\sqrt{2x^2-4x+3}}\\ & =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{(x-1{)}^2+\frac{1}{2}}}\\ \end{array}$

$t=x-1+\sqrt{(x-1{)}^2+\frac{1}{2}}$とおくと$dt=\frac{x-1+\sqrt{(x-1{)}^2+\frac{1}{2}}}{\sqrt{(x-1{)}^2+\frac{1}{2}}}dx$なので、

$\begin{array}{cc}{I}_{31}& =\frac{1}{\sqrt{2}}\int \frac{dt}{t}\\ & =\frac{1}{\sqrt{2}}\log |t|+C\\ & =\frac{1}{\sqrt{2}}\log (x-1+\sqrt{(x-1{)}^2+\frac{1}{2}})+C\\ & =\frac{1}{\sqrt{2}}\log (x-1+\sqrt{x^2-2x+\frac{3}{2}})+C\end{array}$

$\begin{array}{cc}{I}_{32}& =\int \frac{\tan x}{{\cos }^{3}x}dx\\ & =\int \frac{\sin x}{{\cos }^{4}x}dx\\ \end{array}$

$t=\cos x$とおくと$dt=-\sin xdx$なので、

$\begin{array}{cc}{I}_{32}& =-\int \frac{dt}{t^4}\\ & =\frac{1}{3t^3}+C\\ & =\frac{1}{3{\cos }^{3}x}+C\end{array}$

$\begin{array}{cc}{I}_{33}& =\int \frac{dx}{1+\cos x}\\ & =\int \frac{1-\cos x}{1-{\cos }^{2}x}dx\\ & =\int (\frac{1}{{\sin }^{2}x}-\frac{\cos x}{{\sin }^{2}x})dx\\ & =-\frac{1}{\tan x}-\int \frac{\cos x}{{\sin }^{2}x}dx\end{array}$

$t=\sin x$とおくと$dt=\cos xdx$なので、

$\begin{array}{cc}{I}_{33}& =-\frac{1}{\tan x}-\int \frac{dt}{t^2}\\ & =-\frac{1}{\tan x}+\frac{1}{t}+C\\ & =-\frac{1}{\tan x}+\frac{1}{\sin x}+C\end{array}$

別解

$\begin{array}{cc}{I}_{33}& =\int \frac{dx}{1+\cos x}\\ & =\int \frac{dx}{2{\cos }^2\frac{x}{2}}\\ & =\tan \frac{x}{2}+C\end{array}$

$\begin{array}{cc}{I}_{34}& =\int x{\tan }^{2}xdx\\ & =\int (\frac{x}{{\cos }^{2}x}-x)dx\\ & =x\tan x-\int \tan xdx-\frac{1}{2}{x}^2\\ & =x\tan x+\log |\cos x|-\frac{1}{2}{x}^2+C\end{array}$

$\begin{array}{cc}{I}_{35}& =\int e^x\cos xdx\\ & =e^x\cos x+\int e^x\sin xdx\\ & =e^x\cos x+e^x\sin x-\int e^x\cos xdx\\ & =e^x\cos x+e^x\sin x-I_{35}\end{array}$

なので、

$\begin{array}{cc}2I_{35}& =e^x\sin x+e^x\cos x+C^{\prime }\\ I_{35}& =\frac{e^x}{2}(\sin x+\cos x)+C\end{array}$

$\begin{array}{cc}{I}_{36}& =\int \sqrt{x^2-x-1}dx\\ & =\int \sqrt{{(x-\frac{1}{2})}^2-\frac{5}{4}}dx\\ \end{array}$

$t=x-\frac{1}{2}+\sqrt{{(x-\frac{1}{2})}^2-\frac{5}{4}}$とすると$dt=\frac{x-\frac{1}{2}+\sqrt{{(x-\frac{1}{2})}^2-\frac{5}{4}}}{\sqrt{{(x-\frac{1}{2})}^2-\frac{5}{4}}}dx$なので、

$\begin{array}{cc}{I}_{36}& =\int \frac{{(\frac{t}{2}+\frac{5}{8t})}^2-\frac{5}{4}}{t}dt\\ & =\frac{1}{2}\int (\frac{t}{2}-\frac{5}{4t}+\frac{25}{32t^3})dt\\ & =\frac{1}{2}(\frac{t^2}{4}-\frac{5}{4}\log |t|-\frac{25}{64t^2})+C\\ & =\frac{1}{2}((x-\frac{1}{2})\sqrt{x^2-x-1}-\frac{5}{4}\log |x-\frac{1}{2}+\sqrt{x^2-x-1}|)+C\end{array}$

別解

$t=x-\frac{1}{2}+\sqrt{{(x-\frac{1}{2})}^2-\frac{5}{4}}$と置く。…①

${\{t-(x-\frac{1}{2})\}}^2={\left(\sqrt{{(x-\frac{1}{2})}^2-\frac{5}{4}}\right)}^2$

$2t(x-\frac{1}{2})=t^2+\frac{5}{4}$

$x-\frac{1}{2}=\frac{1}{2}(t+\frac{5}{4t})$…②

①②より、$\sqrt{{(x-\frac{1}{2})}^2-\frac{5}{4}}=t-(x-\frac{1}{2})=t-\frac{1}{2}(t+\frac{5}{4t})=\frac{1}{2}(t-\frac{5}{4t})$…③

②をtで微分すると、$\frac{dx}{dt}=\frac{1}{2}-\frac{5}{8t^2}=\frac{1}{2}(1-\frac{5}{4t^2})$

$dx=\frac{1}{2}(1-\frac{5}{4t^2})dt$…④

②③④より

$\begin{array}{cc}{I}_{36}& =\int \frac{1}{2}(t-\frac{5}{4t})\cdot \frac{1}{2}(1-\frac{5}{4t^2})dt\\ & =\frac{1}{4}\int (t-\frac{10}{4t}+\frac{25}{16t^3})dt\\ & =\frac{1}{4}(\frac{t^2}{2}-\frac{5}{2}\log \left|t\right|-\frac{25}{32t^2})+C\\ & =\frac{1}{8}(t^2-\frac{25}{16t^2})-\frac{5}{8}\log \left|t\right|+C\\ & =\frac{1}{2}\cdot \frac{1}{2}(t+\frac{5}{4t})\frac{1}{2}(t-\frac{5}{4t})-\frac{5}{8}\log \left|t\right|+C\\ & =\frac{1}{2}(x-\frac{1}{2})\sqrt{x^2-x-1}-\frac{5}{8}\log |x-\frac{1}{2}+\sqrt{x^2-x-1}|+C\\ \end{array}$

$\begin{array}{cc}{I}_{37}& =\int \sin 2x\cdot \cos 3xdx\\ & =\frac{1}{2}\int (\sin 5x-\sin x)dx\\ & =-\frac{1}{10}\cos 5x+\frac{1}{2}\cos x+C\end{array}$

$\begin{array}{cc}{I}_{38}& =\int {(\sin x+\frac{1}{\sin x})}^{2}dx\\ & =\int ({\sin }^{2}x+2+\frac{1}{{\sin }^{2}x})dx\\ & =\int (\frac{1-\cos 2x}{2}+2+\frac{1}{{\sin }^{2}x})dx\\ & =\frac{5}{2}x-\frac{\sin 2x}{4}-\frac{1}{\tan x}+C\end{array}$

$\begin{array}{cc}{I}_{39}& =\int \frac{1+{\cos }^{3}x}{{\cos }^{2}x}dx\\ & =\int (\frac{1}{{\cos }^{2}x}+\cos x)dx\\ & =\tan x+\sin x+C\end{array}$

$\begin{array}{cc}{I}_{40}& =\int \frac{dx}{1+\sin x}\\ & =\int \frac{1-\sin x}{1-{\sin }^{2}x}dx\\ & =\int (\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x})dx\\ & =\tan x-\int \frac{\sin x}{{\cos }^{2}x}dx\\ \end{array}$

$t=\cos x$とおくと$dt=-\sin xdx$なので、

$\begin{array}{cc}{I}_{40}& =\tan x+\int \frac{dt}{t^2}\\ & =\tan x-\frac{1}{t}+C\\ & =\tan x-\frac{1}{\cos x}+C\end{array}$

$\begin{array}{cc}{I}_{41}& =\int \frac{dx}{e^x+1}\end{array}$

$t=e^x$とおくと$dt=e^{x}dx$なので、

$\begin{array}{cc}{I}_{41}& =\int \frac{dt}{t(t+1)}\\ & =\int (\frac{1}{t}-\frac{1}{t+1})dt\\ & =\log |t|-\log |t+1|+C\\ & =x-\log (e^x+1)+C\end{array}$

$\begin{array}{cc}{I}_{42}& =\int {(\log x)}^{2}dx\\ & =x{(\log x)}^2-2\int \log xdx\\ & =x{(\log x)}^2-2x\log x+2x+C\end{array}$

$\begin{array}{cc}{I}_{43}& =\int \frac{\sqrt{1-x^2}}{x}dx\end{array}$

$t=\sqrt{1-x^2}$とおくと$dt=-\frac{x}{\sqrt{1-x^2}}dx$なので、

$\begin{array}{cc}{I}_{43}& =\int \frac{t^2}{t^2-1}dt\\ & =\int (1+\frac{1}{t^2-1})dt\\ & =\int (1+\frac{1}{2(t-1)}-\frac{1}{2(t+1)})dt\\ & =t+\log \sqrt{\left|\frac{t-1}{t+1}\right|}+C\\ & =\sqrt{1-x^2}+\log \frac{1-\sqrt{1-x^2}}{|x|}+C\end{array}$

$\begin{array}{cc}{I}_{44}& =\int \frac{dx}{x\sqrt{2x-x^2}}\\ \end{array}$

$t=\frac{1}{x}$とおくと$dt=-\frac{1}{x^2}dx$なので、

$\begin{array}{cc}{I}_{44}& =-\int \frac{dt}{\sqrt{2t-1}}\\ & =-\sqrt{2t-1}+C\\ & =-\sqrt{\frac{2-x}{x}}+C\end{array}$