integral from 0 to sin(x) of ln(t^3+1)

Lösung

\int_{0}^{s i n (x)} ln (t^{3} + 1) d t

sin (x) ln (sin^{3} (x) + 1) + ln ∣ sin (x) + 1 ∣ + \frac{- ln ( \frac{∣ 4 s i n ^{2} ( x ) - 4 s i n ( x ) + 4 ∣}{3} ) + 2 3 arctan ( \frac{2 s i n ( x ) - 1}{3} )}{2} - 3 sin (x) - \frac{- 3 ln ( \frac{4}{3} ) - π}{2 3}

Schritte zur Lösung

\int_{0}^{s i n (x)} ln (t^{3} + 1) d t

Wende die partielle Integration an

= [t ln (t^{3} + 1) - \int \frac{3 t ^{3}}{t ^{3} + 1} d t]_{0}^{s i n (x)}

\int \frac{3 t ^{3}}{t ^{3} + 1} d t = - ln ∣ t + 1 ∣ - \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) + 3 t

= [t ln (t^{3} + 1) - (- ln ∣ t + 1 ∣ - \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) + 3 t)]_{0}^{s i n (x)}

Vereinfache

[t ln (t^{3} + 1) - (- ln ∣ t + 1 ∣ - \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) + 3 t)]_{0}^{s i n (x)} : [t ln (t^{3} + 1) + ln ∣ t + 1 ∣ + \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) - 3 t]_{0}^{s i n (x)}

= [t ln (t^{3} + 1) + ln ∣ t + 1 ∣ + \frac{1}{2} (- ln \frac{4 t ^{2} - 4 t}{3} + \frac{4}{3} + 23 arctan (\frac{2 t - 1}{3})) - 3 t]_{0}^{s i n (x)}

Berechne die Grenzen:

sin (x) ln (sin^{3} (x) + 1) + ln ∣ sin (x) + 1 ∣ + \frac{- ln \frac{4 s i n ^{2} ( x ) - 4 s i n ( x )}{3} + \frac{4}{3} + 2 3 arctan ( \frac{2 s i n ( x ) - 1}{3} )}{2} - 3 sin (x) - \frac{- 3 ln ( \frac{4}{3} ) - π}{2 3}

= sin (x) ln (sin^{3} (x) + 1) + ln ∣ sin (x) + 1 ∣ + \frac{- ln \frac{4 s i n ^{2} ( x ) - 4 s i n ( x )}{3} + \frac{4}{3} + 2 3 arctan ( \frac{2 s i n ( x ) - 1}{3} )}{2} - 3 sin (x) - \frac{- 3 ln ( \frac{4}{3} ) - π}{2 3}

Vereinfache

= sin (x) ln (sin^{3} (x) + 1) + ln ∣ sin (x) + 1 ∣ + \frac{- ln ( \frac{∣ 4 s i n ^{2} ( x ) - 4 s i n ( x ) + 4 ∣}{3} ) + 2 3 arctan ( \frac{2 s i n ( x ) - 1}{3} )}{2} - 3 sin (x) - \frac{- 3 ln ( \frac{4}{3} ) - π}{2 3}

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