substitutionsverfahren

integrationsmethode-lineare-substitution

Integratinsmethode - lineare Substitution $\int f(a\ast x+b)dx=\frac{F(a\ast x+b)}{a}+c$ innere Funktion ist linear ⇒ u = ax + b 6.37) a) mit Formel: $u=1-3x=ax+b$ $a=-3$ $b=1$ $f(x)=x^3\to F(x)=\frac{x^4}{4}$ $\int (1-3x{)}^{3}dx=\frac{(1-3x{)}^4}{4}\ast \frac{1}{-3}+c=-\frac{1}{12}\ast (1-3x{)}^4+c$

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Substitutionsverfahren

$\int e^{\sin x}\cos xdx$ $=\int e^u=\int e^{u}du=e^u+e=e^{\sin x}+c$

Substituieren

$u=\sin x$ $\frac{du}{dx}=\cos x$ $dx=\frac{dx}{\cos x}$

Probe

$(e^{\sin x}+c{)}^{\prime }=e^{\sin x}\ast \cos x$ Bsp: $\int \sin x^3\ast x^{2}dx$ $=\int \frac{1}{3}\sin (u)du=-\frac{1}{3}\cos u+c$ $=-\frac{1}{3}\cos (x^3)+c$ $u=x^3$ $\frac{du}{dx}=3x^2$ $dx=\frac{du}{2x^2}$

6.45) a) $\int \frac{x}{(x^2+1{)}^2}dx;u=x^2+1$ $\frac{du}{dx}=2x\Rightarrow dx=\frac{1}{2x}\ast du$ $\int \frac{x}{u^2}\ast \frac{1}{2x}du=\int \frac{1}{u^2}\ast \frac{1}{2}du=\frac{1}{2}\ast \int u^{-2}du=\frac{1}{2}\ast \frac{u^{-1}}{-1}+c$ $\int \frac{x}{(x^1+1{)}^2}dx=-\frac{1}{2}\ast \frac{1}{x^2+1}+c$ c) $\int \frac{\cos x}{{\sin }^{2}x}dx;u=\sin x$ $dx=\frac{du}{\cos x}$ $\int \frac{\cos x}{u^2}\ast \frac{1}{\cos x}du=\int \frac{1}{u^2}du=\int u^{-2}\ast$ $\frac{\cos x}{{\sin }^{2}x}\ast dx=-\frac{1}{\sin (x)}+c$

lineare-substitution-und-integrale

Bsp:Lineares Substituioren $\int \sin (3x-1)dx$ $=\int \sin u\ast x^2\ast \frac{du}{3x^2}$ $\int \frac{1}{3}\sin udu=-\frac{1}{3}\cos u+c$ $=-\frac{1}{3}\cos (x^3)+c$ $u=3x-1$ $\frac{du}{dx}=3$ $dx=\frac{du}{3}$

6.38) a) $\int \frac{2}{2+x}dx$

6.37) b) $\int (3-2x{)}^{4}dx$ c) $\int \sqrt{1-x}dx$ 6.45) a) $\int \frac{x}{(x^2-1{)}^2}dx;u=x^2+1$ c) $\int \frac{\cos x}{{\sin }^{2}x}dx;u=\sin x$ 6.37) d) $\int \frac{1}{\sqrt{1-x}}dx=\int (1-x{)}^{-1/2}dx$ $u=1-x$ → ableiten $\frac{du}{dx}=-1\Rightarrow dex=\frac{du}{-1}=dx=-du$ $\int u^{-1/2}(-du)=-2\ast u^{1/2}+c=-2\ast \sqrt{1-x}+c$ e) $\int (\frac{x}{2}-2{)}^{2}dx$ $u=\frac{x}{2}-2$ $\frac{du}{dx}=\frac{1}{2}$ $du=\frac{1}{2}\ast dx|:\frac{1}{2}|\ast 2$ $dx=2\ast du$ $\int u^2\ast 2\ast du=2\ast \int u^2\ast du=2\ast \frac{u^3}{3}+c$ $\frac{2}{3}\ast (\frac{x}{2}-2{)}^3+c$

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linerare-substitutions-aufgaben

6.42)

1. unbestimmtes Integral lösen (Grenzen ignoreiren)
2. Grezen einsetzen a) ${\int }_0^4(\frac{x}{3}-3{)}^{2}dx$ c) ${\int }_1^{2}2\ast \frac{1}{3-x}dx$ 6.43) a) ${\int }_0^1{e}^{4x-2}dx=$ $u=4x-2$ $\frac{du}{dx}=4$ $dx=\frac{1}{4}du$ $\int e^u\ast \frac{1}{4}du$ $\frac{1}{4}\int e^{u}du=\frac{e^u}{4}+c$ grenzen einsetzen $=\frac{e^{4\ast 1\ast 2}}{4}-\frac{e^{4\ast 0\ast 2}}{4}\approx 1.81$ c) ${\int }_0^{\pi /4}\sin 2xdx$ $u=2x$ $\frac{du}{dx}=2\Rightarrow \frac{1}{2}\ast du=dx$ $\int \sin u\ast \frac{1}{2}du=\frac{1}{2}\ast (-\cos u)+c$ $\frac{1}{2}\ast (-\cos 2x)+c$ $\frac{-\cos 2x}{2}{|}_0^{\pi /4}=\frac{-\cos \frac{\pi }{2}}{2}-\frac{-\cos 0}{2}=0.5$

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