{"id":4262,"date":"2013-12-11T06:05:05","date_gmt":"2013-12-10T21:05:05","guid":{"rendered":"http:\/\/fujiitoshiki.com\/improvesociety\/?p=4262"},"modified":"2014-08-01T19:23:53","modified_gmt":"2014-08-01T10:23:53","slug":"integrals","status":"publish","type":"post","link":"https:\/\/www.fujiitoshiki.com\/improvesociety\/?p=4262","title":{"rendered":"Integrals"},"content":{"rendered":"
\n

Integrals<\/h3>\n

If dy\/dx = f(x), then it’s called that y an indefinite integral of f(x) and denoted as<\/p>\n

\\displaystyle \\int f(x)dx<\/p>\n

If f(x) = \\frac{d}{dx}F(x), then<\/p>\n

\\displaystyle \\int_a^b f(x)dx = F(b) - F(a)<\/p>\n

Integral formulas<\/h3>\n

In the following u, v represent functions of x while a, b, c, p represent constants. <\/p>\n

\\displaystyle \\int (u \\pm v)dx = \\int u dx \\pm \\int v dx\\\\\\vspace{0.2 in} \\int cu dx = c\\int u dx \\\\\\vspace{0.2 in} \\int u\\left(\\frac{dv}{dx}\\right) = uv - \\int v \\left(\\frac{du}{dx}\\right)dx \\\\\\vspace{0.2 in} \\int u dv = uv - \\int v du <\/p>\n

This is called integration by parts.<\/p>\n

\\displaystyle \\int F(u(x))dx = \\int F(w)\\frac{dw}{dw\/dx}<\/p>\n

where w = u(x) and w’ = dw\/dx expressed as a function of w. This is called integration by substitution or tranformation.<\/p>\n

\\displaystyle \\int u^p du = \\frac{u^{p+1}}{p+1},\\ p \\neq -1\\\\\\vspace{0.2 in} \\int u^{-1}du = \\int \\frac{du}{u} = \\ln u\\\\\\vspace{0.2 in} \\int a^u du = \\frac{a^u}{\\ln a},\\ a \\neq 0,\\ 1\\\\\\vspace{0.2 in} \\int e^u du = e^u<\/p>\n<\/p>\n

\\displaystyle \\int \\sin u\\ du = -\\cos{u}\\\\\\vspace{0.2 in} \\int \\cos u\\ du = \\sin{u}\\\\\\vspace{0.2 in} \\int \\tan u\\ du = -\\ln \\cos{u}\\\\\\vspace{0.2 in} \\int e^{au}\\sin{bu}\\ du = \\frac{e^{au}(a\\ \\sin{bu}- b\\ \\cos{bu})}{a^2 + b^2}\\\\\\vspace{0.2 in} \\int e^{au}\\cos{bu}\\ du = \\frac{e^{au}(a\\ \\cos{bu}+ b\\ \\sin{bu})}{a^2 + b^2}\\\\\\vspace{0.2 in} \\int \\frac{du}{\\sqrt{a^2 - u^2}} = \\sin^{-1}\\frac{u}{a}\\\\\\vspace{0.2 in} \\int \\frac{du}{u^2 + a^2} = \\frac{1}{a}\\tan^{-1}\\frac{u}{a}\\\\\\vspace{0.2 in} \\int \\frac{du}{\\sqrt{u^2 - a^2}} = \\ln(u + \\sqrt{u^2 - a^2})\\\\\\vspace{0.2 in} \\int \\frac{du}{\\sqrt{u^2 + a^2}} = \\ln(u + \\sqrt{u^2 + a^2})<\/p>\n<\/blockquote>\n