数学の基礎と公式-7

数学の基礎と公式

amazon kindle版を出版しました。

---

7. 不定積分-3　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)

$\int \sin x \cos x dx=-\displaystyle{{{\cos ^2x}\over{2}}}$

$\int (\sin x \cos x)^2 dx=\displaystyle{-{{\sin \left(4\,x\right)-4\,x}\over{32}}}$

$\int (\sin x \cos x)^3 dx=\displaystyle{-{{2\,\sin ^6x-3\,\sin ^4x}\over{12}}}$

$\int (\sin x \cos x)^4 dx=\displaystyle{{{\sin \left(8\,x\right)-8\,\sin \left(4\,x\right)+24\,x}\over{1024 }}}$

$\int (\sin x \cos x)^5 dx=\displaystyle{{{6\,\sin ^{10}x-15\,\sin ^8x+10\,\sin ^6x}\over{60}}}$

(2)

$\int (\sin x)^2 \cos x dx=\displaystyle{{{\sin ^3x}\over{3}}}$

$\int (\sin x)^3 \cos x dx=\displaystyle{{{\sin ^4x}\over{4}}}$

$\vdots$

$\int (\sin x)^n \cos x dx=\displaystyle{{{\sin ^{n+1}x}\over{n+1}}}$

(3)

$\int \sin x (\cos x)^2 dx=-\displaystyle{{{\cos ^3x}\over{3}}}$

$\int \sin x (\cos x)^3 dx=-\displaystyle{{{\cos ^4x}\over{4}}}$

$\vdots$

$\int \sin x (\cos x)^n dx=\displaystyle{{{\cos ^{n+1}x}\over{n+1}}}$

(4)$\displaystyle{}$

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

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

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

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

$\displaystyle{\int \dfrac{1}{\sin^5 x \cos x} dx=-{{\log \left(\sin ^2x-1\right)}\over{2}}+\log \sin x-{{2\,\sin ^2x +1}\over{4\,\sin ^4x}}}$

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

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

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

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

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

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

$\displaystyle{\int \dfrac{1}{(\sin x \cos x)^4} dx={{\tan ^3x+9\,\tan x}\over{3}}-{{9\,\tan ^2x+1}\over{3\,\tan ^3x}}}$

$\displaystyle{\int \dfrac{1}{(\sin x \cos x)^5}} dx$

$\displaystyle{=-3\,\log \left(\sin ^2x-1\right)+6\,\log \sin x-{{12\,\sin ^6x-18\, \sin ^4x+4\,\sin ^2x+1}\over{4\,\sin ^8x-8\,\sin ^6x+4\,\sin ^4x}}}$