数学の基礎と公式-6 - 橋平礼の電験三種合格講座

数学の基礎と公式

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6. 不定積分-2　三角関数と双曲線関数(積分定数Cは省略しています。)

※$\log x$の$x$の箇所には絶対値がつきます。$\log | x |$

(1)$\int \sin x dx=-\cos x,~~\int \cos x dx=\sin x,~~\displaystyle \int\dfrac{1}{\cos^2 x} dx=\tan x$

(2)

$\int \sin 2 x dx=-\dfrac{\cos 2 x}{2}$

$\int \sin 3 x dx=-\dfrac{\cos 3 x}{3}$

$ \vdots$

$\int \sin n x dx=-\dfrac{\cos n x}{n}$

$\int (\sin x)^2 dx=-\dfrac{\sin 2 x-2 x}{4}$

$\int (\sin x)^3 dx=\dfrac{\cos^3 x-3 \cos x}{3}$

$\int (\sin x)^4 dx={\dfrac{\sin \left(4\,x\right)-8\,\sin \left(2\,x\right)+12\,x}{32}}$

$\int (\sin x)^5 dx=-{\dfrac{3\,\cos ^5x-10\,\cos ^3x+15\,\cos x}{15}}$

(3)

$\int (\cos x)^2 dx={\dfrac{\sin \left(2\,x\right)+2\,x}{4}}$

$\int (\cos x)^3 dx=-{\dfrac{\sin ^3x-3\,\sin x}{3}}$

$\int (\cos x)^4 dx={\dfrac{\sin \left(4\,x\right)+8\,\sin \left(2\,x\right)+12\,x}{32}}$

$\int (\cos x)^5 dx={\dfrac{3\,\sin ^5x-10\,\sin ^3x+15\,\sin x}{15}}$

(4)

$\int (\tan x)^2 dx=\tan x-x$

$\int (\tan x)^3 dx={\dfrac{\log \left(\sin ^2x-1\right)}{2}}-{\dfrac{1}{2\,\sin ^2x-2}}$

$\int (\tan x)^4 dx={\dfrac{\tan ^3x-3\,\tan x}{3}}+x$

$\int (\tan x)^5 dx={\dfrac{4\,\sin ^2x-3}{4\,\sin ^4x-8\,\sin ^2x+4}}-{\dfrac{\log \left( \sin ^2x-1\right)}{2}}$

(5)

$\int cosec \, x dx=-\log \left(cosec \,x+\cot x\right)$

$\int (cosec \, x)^2 dx=-{\dfrac{1}{\tan x}}$

$\int (cosec \, x)^3 dx=-{\dfrac{\log \left(\cos x+1\right)}{4}}+{\dfrac{\log \left(\cos x-1 \right)}{4}}+{\dfrac{\cos x}{2\,\cos ^2x-2}}$

$\int (cosec \, x)^4 dx=-{\dfrac{3\,\tan ^2x+1}{3\,\tan ^3x}}$

$\int (cosec \, x)^5 dx=-{\dfrac{3\,\log \left(\cos x+1\right)}{16}}+{\dfrac{3\,\log \left(\cos x- 1\right)}{16}}+{\dfrac{3\,\cos ^3x-5\,\cos x}{8\,\cos ^4x-16\, \cos ^2x+8}}$

(6)

$\int \sec x dx=\log \left(\tan x+\sec x\right)$

$\int (\sec x)^2 dx=\tan x$

$\int (\sec x)^3 dx={\dfrac{\log \left(\sin x+1\right)}{4}}-{\dfrac{\log \left(\sin x-1\right) }{4}}-{\dfrac{\sin x}{2\,\sin ^2x-2}}$

$\int (\sec x)^4 dx={\dfrac{\tan ^3x}{3}}+\tan x$

$\int (\sec x)^5 dx={\dfrac{3\,\log \left(\sin x+1\right)}{16}}-{\dfrac{3\,\log \left(\sin x-1 \right)}{16}}-{\dfrac{3\,\sin ^3x-5\,\sin x}{8\,\sin ^4x-16\, \sin ^2x+8}}$

(7)

$\int \cot x dx=\log( \sin x)$

$\int (\cot x)^2 dx=-{\dfrac{1}{\tan x}}-x$

$\int (\cot x)^3 dx=-\log (\sin x)-{\dfrac{1}{2\,\sin ^2x}}$

$\int (\cot x)^4 dx={\dfrac{3\,\tan ^2x-1}{3\,\tan ^3x}}+x$

$\int (\cot x)^5 dx=\log (\sin x)+{\dfrac{4\,\sin ^2x-1}{4\,\sin ^4x}}$

(8)

$\int \arcsin x dx=x\, \arcsin x+\sqrt{1-x^2}$

$\int (\arcsin x)^2 dx=x\,\arcsin ^2x+2\,\sqrt{1-x^2}\,\arcsin x-2\,x$

$\int (\arcsin x)^3 dx=x\,\arcsin ^3x+3\,\sqrt{1-x^2}\,\arcsin ^2x-6\,\left(x\,\arcsin x+ \sqrt{1-x^2}\right)$

$\int (\arcsin x)^4 dx=x\,\arcsin ^4x+\sqrt{1-x^2}\,\left(4\,\arcsin ^3x-24\,\arcsin x \right)-12\,x\,\arcsin ^2x+24\,x$

$\int (\arcsin x)^5 dx=x\,\arcsin ^5x+\sqrt{1-x^2}\,\left(5\,\arcsin ^4x-60\,\arcsin ^2x+120\right)-20\,x\,\arcsin ^3x+120\,x\,\arcsin x$

(9)

$\int \arctan x dx=x\,\arctan x-\displaystyle {{\log \left(x^2+1\right)}\over{2}}$

(10) $\int \sinh x dx=\cosh x,~~\int \cosh x dx=\sinh x,~~ \int\tanh x dx=\log(\cosh x)$

(11)

$\int (\sinh x)^2 dx=\displaystyle {{e^{2\,x}}\over{8}}-{{e^ {- 2\,x }}\over{8}}-{{x}\over{2}}$

$\int (\sinh x)^3 dx=\displaystyle {{{e^{3\,x}}\over{24}}-{{3\,e^{x}}\over{8}}-{{3\,e^ {- x }}\over{8}} +{{e^ {- 3\,x }}\over{24}}}$

$\int (\sinh x)^4 dx=\displaystyle {{{e^{4\,x}}\over{64}}-{{e^{2\,x}}\over{8}}+{{e^ {- 2\,x }}\over{8}} -{{e^ {- 4\,x }}\over{64}}+{{3\,x}\over{8}}}$

$\int (\sinh x)^5 dx=\displaystyle {{{e^{5\,x}}\over{160}}-{{5\,e^{3\,x}}\over{96}}+{{5\,e^{x}}\over{16 }}+{{5\,e^ {- x }}\over{16}}-{{5\,e^ {- 3\,x }}\over{96}}+{{e^ {- 5 \,x }}\over{160}}}$

(12)

$\int (\cosh x)^2 dx=\displaystyle {{e^{2\,x}}\over{8}}-{{e^ {- 2\,x }}\over{8}}+{{x}\over{2}}$

$\int (\cosh x)^3 dx=\displaystyle {{{e^{3\,x}}\over{24}}+{{3\,e^{x}}\over{8}}-{{3\,e^ {- x }}\over{8}} -{{e^ {- 3\,x }}\over{24}}}$

$\int (\cosh x)^4 dx=\displaystyle {{{e^{4\,x}}\over{64}}+{{e^{2\,x}}\over{8}}-{{e^ {- 2\,x }}\over{8}} -{{e^ {- 4\,x }}\over{64}}+{{3\,x}\over{8}}}$

$\int (\cosh x)^5 dx=\displaystyle {{{e^{5\,x}}\over{160}}+{{5\,e^{3\,x}}\over{96}}+{{5\,e^{x}}\over{16 }}-{{5\,e^ {- x }}\over{16}}-{{5\,e^ {- 3\,x }}\over{96}}-{{e^ {- 5 \,x }}\over{160}}}$

(13)

$\int (\tanh x)^2 dx=\displaystyle {x-{{2}\over{e^ {- 2\,x }+1}}}$

$\int (\tanh x)^3 dx=\displaystyle {\log \left(e^ {- 2\,x }+1\right)+{{2\,e^ {- 2\,x }}\over{2\,e^ {- 2 \,x }+e^ {- 4\,x }+1}}+x}$

$\int (\tanh x)^4 dx=\displaystyle {x-{{12\,e^ {- 2\,x }+12\,e^ {- 4\,x }+8}\over{9\,e^ {- 2\,x }+9\,e ^ {- 4\,x }+3\,e^ {- 6\,x }+3}}}$

$\int (\tanh x)^5 dx=\displaystyle {\log \left(e^ {- 2\,x }+1\right)+{{4\,e^ {- 2\,x }+4\,e^ {- 4\,x }+ 4\,e^ {- 6\,x }}\over{4\,e^ {- 2\,x }+6\,e^ {- 4\,x }+4\,e^ {- 6\,x }+e^ {- 8\,x }+1}}+x}$