integral from 0 to sin(pi) of ln(2+t^2)

Lösung

\int_{0}^{s i n (π)} ln (2 + t^{2}) d t

0

Schritte zur Lösung

\int_{0}^{s i n (π)} ln (2 + t^{2}) d t

Wende die partielle Integration an

= [t ln (2 + t^{2}) - \int \frac{2 t ^{2}}{2 + t ^{2}} d t]_{0}^{0}

\int \frac{2 t ^{2}}{2 + t ^{2}} d t = - 22 arctan (\frac{t}{2}) + 2 t

= [t ln (2 + t^{2}) - (- 22 arctan (\frac{t}{2}) + 2 t)]_{0}^{0}

Vereinfache

= [t ln (2 + t^{2}) + 22 arctan (\frac{t}{2}) - 2 t]_{0}^{0}

Berechne die Grenzen:

0

= 0

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