数学の基礎と公式-8 - 橋平礼の電験三種合格講座

数学の基礎と公式

amazon kindle版を出版しました。

8. 不定積分-4　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)$\displaystyle{}$

$\displaystyle{\int \dfrac{1}{1+\sin x} dx=-{{2\,\cos x+2}\over{\sin x+\cos x+1}}}$

$\displaystyle{\int \dfrac{1}{1+\sin^2 x} dx={{\arctan \left(\sqrt{2}\,\tan x\right)}\over{\sqrt{2}}}}$

$\displaystyle{\int \dfrac{1}{1-\sin x} dx=-{{2\,\cos x+2}\over{\sin x-\cos x-1}}}$

$\displaystyle{\int \dfrac{1}{1-\sin^2 x} dx=\tan x}$

$\displaystyle{\int \dfrac{1}{1+\cos x} dx={{\sin x}\over{\cos x+1}}}$

$\displaystyle{\int \dfrac{1}{1+\cos^2 x} dx={{\arctan \left({{\tan x}\over{\sqrt{2}}}\right)}\over{\sqrt{2}}}}$

$\displaystyle{\int \dfrac{1}{1-\cos x} dx=-{{\cos x+1}\over{\sin x}}}$

$\displaystyle{\int \dfrac{1}{1-\cos^2 x} dx=-{{1}\over{\tan x}}}$

(2)

$\displaystyle{\int x \sin x dx=\sin x-x\,\cos x}$

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

$\displaystyle{\int x^3 \sin x dx=\left(3\,x^2-6\right)\,\sin x+\left(6\,x-x^3\right)\,\cos x}$

$\displaystyle{\int x \cos x dx=x\,\sin x+\cos x}$

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

$\displaystyle{\int x^3 \cos x dx=\left(x^3-6\,x\right)\,\sin x+\left(3\,x^2-6\right)\,\cos x}$

(3)

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

$\displaystyle{\int x \arccos x dx=-{{-\arcsin x-2\,x^2\,\arccos x+x\,\sqrt{1-x^2}}\over{4}}}$

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

$\displaystyle{\int x \,{\rm arccsc} \,x dx={{x^2\,{\rm arccsc}\; x\, +\sqrt{x^2-1}}\over{2 }}}$

$\displaystyle{\int x \,{\rm arccot} \,x dx=-{{\arctan x-x^2\,{\rm arccot}\; x-x}\over{2}}}$

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

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

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

(4)

$\displaystyle{e^x \,\sin x dx={{e^{x}\,\left(\sin x-\cos x\right)}\over{2}}}$

$\displaystyle{e^x \,\cos x dx={{e^{x}\,\left(\sin x+\cos x\right)}\over{2}}}$

$\displaystyle{e^ax \,\sin bx dx={{\left(a\,\sin \left(b\,x\right)-b\,\cos \left(b\,x\right)\right) \,e^{a\,x}}\over{a^2+b^2}}}$

$\displaystyle{e^ax \,\cos bx dx={{\left(a\,\cos \left(b\,x\right)+b\,\sin \left(b\,x\right)\right) \,e^{a\,x}}\over{a^2+b^2}}}$

$\displaystyle{e^{ax} \,(\sin x)^2 dx=-{{\left(\left(\cos \left(2\,x\right)-1\right)\,a^2+2\,\sin \left(2 \,x\right)\,a-4\right)\,e^{a x}}\over{2\,a^3+8\,a}}}$

$\displaystyle{e^{ax} \,(\cos x)^2 dx={{\left(\left(\cos \left(2\,x\right)+1\right)\,a^2+2\,\sin \left(2 \,x\right)\,a+4\right)\,e^{x\,a}}\over{2\,a^3+8\,a}}}$

(5)

$\displaystyle{\sin(\log x) dx={{x\,\left(\sin \log x-\cos \log x\right)}\over{2}}}$

$\displaystyle{\cos(\log x) dx={{x\,\left(\sin \log x+\cos \log x\right)}\over{2}}}$