substitutionsverfahren

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

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