=for timestamp
Mi Nov 22 18:11:36 CET 2006

=head2 114. Hausaufgabe

=head3 Analysis-Buch Seite 256, Aufgabe 14a

M<\int e^x \sin x \,\mathrm{d}x =
{}e^x \sin x - \int e^x \cos x \,\mathrm{d}x =
{}e^x \sin x - e^x \cos x - \int e^x \sin x \,\mathrm{d}x;> ⇔
M<\int e^x \sin x \,\mathrm{d}x = \frac{e^x}{2}\left(\sin x - \cos x\right)\!;>

M<\int\limits_0^{\pi/2} e^x \sin x \,\mathrm{d}x =
{}\frac{1}{2}\left(e^{\pi/2} + 1\right)\!;>

=head3 Analysis-Buch Seite 256, Aufgabe 15

=over

=item e)

M<\displaystyle\int\limits_{-1}^1 \ln x^2 \,\mathrm{d}x =
{}2 \int\limits_0^1 \ln x^2 \,\mathrm{d}x =
{}2 \int\limits_0^1 x' \cdot \ln x^2 \,\mathrm{d}x =
{}\lim\limits_{\alpha \to 0+} 2 \left[x \ln x^2 - \int \underbrace{x \cdot \frac{1}{x^2} \cdot 2x}_{2} \,\mathrm{d}x\right]_{\alpha}^1 =
{}\lim\limits_{\alpha \to 0+} 2 \left[x \ln x^2 - 2x\right]_{\alpha}^1 = -4;>

=item f)

=for latex
{\footnotesize

M<\displaystyle\int\limits_1^{\sqrt{2}} x \ln\!\left(1+x^2\right) \,\mathrm{d}x =
{}\int\limits_1^{\sqrt{2}} \left(\frac{1}{2} x^2\right)' \ln\!\left(1+x^2\right) \,\mathrm{d}x =
{}\left[\frac{1}{2} x^2 \ln\!\left(1 + x^2\right) - \int \underbrace{\frac{1}{2} x^2 \cdot \frac{1}{1 + x^2} \cdot 2x}_{\frac{x^3}{1+x^2}} \,\mathrm{d}x\right]_1^{\sqrt{2}} =
{}\left[\frac{1}{2} x^2 \ln\!\left(1 + x^2\right) - \int x \,\mathrm{d}x - \underbrace{\frac{x}{1 + x^2}}_{\frac{\left(1 + x^2\right)'}{1 + x^2} \cdot \frac{1}{2}} \mathrm{d}x\right]_1^{\sqrt{2}} =
{}\left[\frac{1}{2} x^2 \ln\!\left(1 + x^2\right) - \frac{1}{2} x^2 + \frac{1}{2} \ln \left|1 + x^2\right|\right]_1^{\sqrt{2}} =
{}\frac{1}{2} \left[\left(1 + x^2\right) \ln\!\left(1 + x^2\right) - x^2\right]_1^{\sqrt{2}} =
{}\frac{1}{2}\left(3 \ln 3 - 2 \ln 2 - 1\right);>

=for latex
}

=item g)

M<\displaystyle\int\limits_0^1 x^{-1/2} \ln x \,\mathrm{d}x =
{}\int\limits_0^1 \left(2 x^{1/2}\right)' \ln x \,\mathrm{d}x =
{}\lim\limits_{\alpha \to 0+} \left[2 \sqrt{x} \ln x - \int 2 \underbrace{x^{1/2} x^{-1}}_{x^{-1/2}} \,\mathrm{d}x\right]_{\alpha}^1 =
{}\lim\limits_{\alpha \to 0+} \left[2 \sqrt{x} \left(\ln x - 2\right)\right]_{\alpha}^1 =
{}-4;>

=back

=head3 Analysis-Buch Seite 256, Aufgabe 17a

Zeige, dass gilt:

M<\displaystyle\int \sin^n x \,\mathrm{d}x =
{}-\frac{1}{n} \left(\sin x\right)^{n-1} \cos x + \frac{n - 1}{n} \int \left(\sin x\right)^{n-2} \mathrm{d}x;>

M<\displaystyle\renewcommand{\arraystretch}{2.2}\begin{array}{@{}cl}
{} & \left[-\frac{1}{n} \left(\sin x\right)^{n-1} \cos x + \frac{n - 1}{n} \int \left(\sin x\right)^{n-2} \mathrm{d}x\right]' = \\
{} =& \frac{1}{n} \left[\left(\sin x\right)^{n-1} \sin x - \left(n-1\right) \left(\sin x\right)^{n-2} \underbrace{\cos^2 x}_{1 - \sin^2 x} + \left(n-1\right) \left(\sin x\right)^{n-2}\right] = \\
{} =& \frac{1}{n} \sin^n x \left[1 - \left(n-1\right) \left(\sin x\right)^{-2} \left(1 - \sin^2 x - 1\right)\right] = \\
{} =& \frac{1}{n} \sin^n x \cdot \left(1 + n - 1\right) = \sin^n x;
\end{array}>