integral from 0 to pi/3 of xtan^2(x)

Lösung

\int_{0}^{\frac{π}{3}} x tan^{2} (x) d x

\frac{π}{3} - \frac{π ^{2}}{18} - ln (2)

Dezimale

0.57234 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{3}} x tan^{2} (x) d x

Wende die partielle Integration an

= [x (- x + tan (x)) - \int - x + tan (x) d x]_{0}^{\frac{π}{3}}

\int - x + tan (x) d x = - \frac{x ^{2}}{2} - ln ∣ cos (x) ∣

= [x (- x + tan (x)) - (- \frac{x ^{2}}{2} - ln ∣ cos (x) ∣)]_{0}^{\frac{π}{3}}

Vereinfache

[x (- x + tan (x)) - (- \frac{x ^{2}}{2} - ln ∣ cos (x) ∣)]_{0}^{\frac{π}{3}} : [\frac{1}{2} (- x^{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣)]_{0}^{\frac{π}{3}}

= [\frac{1}{2} (- x^{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣)]_{0}^{\frac{π}{3}}

Berechne die Grenzen:

\frac{π}{3} - \frac{π ^{2}}{18} - ln (2)

= \frac{π}{3} - \frac{π ^{2}}{18} - ln (2)

t an g e n t 5 x + 4 cos (x)

in v erse l a pl a ce \frac{s}{( s ^{2} + 4 ) ^{2}}

t an g e n t f (x) = 6 sin (x) + 4 cos (x), at x = 0

\int \frac{( arctan ( x ) ) ^{7}}{1 + x ^{2}} d x

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