=for timestamp
Mo Nov 20 16:50:44 CET 2006

=head2 113. Hausaufgabe

=head3 Analysis-Buch Seite 256, Aufgabe 15

=over

=item a)

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

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

M<\int\limits_0^{\pi} \sin^2 x \,\mathrm{d}x =
{}\frac{1}{2} \left[x - \sin x \cos x\right]_0^{\pi} =
{}\frac{\pi}{2};>

=item b)

M<\int\limits_1^e x \ln x \,\mathrm{d}x =
{}\int\limits_1^e \left(\frac{1}{2} x^2\right)' \cdot \ln x \,\mathrm{d}x =
{}\left[\frac{1}{2} x^2 \cdot \ln x - \int \frac{1}{x} \cdot \frac{1}{2} x^2 \,\mathrm{d}x\right]_1^e =
{}\left[\frac{1}{2} x^2 \cdot \ln x - \frac{1}{4} x^2\right]_1^e =
{}\frac{e^2}{4} + \frac{1}{4};>

=item c)

M<\int\limits_1^{e^2} \sqrt{x} \ln x \,\mathrm{d}x =
{}\int\limits_1^{e^2} \left(\frac{2}{3} x^{3/2}\right)' \cdot \ln x \,\mathrm{d}x =
{}\left[\frac{2}{3} x^{3/2} \cdot \ln x - \int \frac{1}{x} \cdot \frac{2}{3} x^{3/2} \,\mathrm{d}x\right]_1^{e^2} =
{}\frac{2}{3} \left[x^{3/2} \cdot \ln x - \frac{2}{3} x^{3/2}\right]_1^{e^2} =
{}\frac{8}{9} e^3 + \frac{4}{9};>

=item d)

=for latex
{\footnotesize

M<\int\limits_{\sqrt{e}}^e \ln^2 x \,\mathrm{d}x =
{}\int\limits_{\sqrt{e}}^e x' \cdot \ln^2 x \,\mathrm{d}x =
{}\left[x \cdot \ln^2 x - \int x \cdot 2 \ln x \cdot \frac{1}{x} \,\mathrm{d}x\right]_{\sqrt{e}}^e =
{}\left[x \cdot \ln^2 x - 2 \int x \ln x \,\mathrm{d}x\right]_{\sqrt{e}}^e =
{}\left[x \cdot \ln^2 x - 2 \left(x \cdot \ln x - x\right)\right]_{\sqrt{e}}^e =
{}\left[x \cdot \left(\ln^2 x - 2 \ln x + 2\right)\right]_{\sqrt{e}}^e =
{}e - \frac{5}{4} \sqrt{e};>

=for latex
}

=back