integral from 1 to e^pi of sin(ln(x))

Lösung

\int_{1}^{e^{π}} sin (ln (x)) d x

\frac{e ^{π} + 1}{2}

Dezimale

12.07034 \dots

Schritte zur Lösung

\int_{1}^{e^{π}} sin (ln (x)) d x

Wende die partielle Integration an

= [x sin (ln (x)) - \int cos (ln (x)) d x]_{1}^{e^{π}}

\int cos (ln (x)) d x = \frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))

= [x sin (ln (x)) - (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))]_{1}^{e^{π}}

Vereinfache

[x sin (ln (x)) - (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))]_{1}^{e^{π}} : [\frac{1}{2} (x sin (ln (x)) - x cos (ln (x)))]_{1}^{e^{π}}

= [\frac{1}{2} (x sin (ln (x)) - x cos (ln (x)))]_{1}^{e^{π}}

Berechne die Grenzen:

\frac{e ^{π}}{2} + \frac{1}{2}

= \frac{e ^{π}}{2} + \frac{1}{2}

Vereinfache

= \frac{e ^{π} + 1}{2}

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