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

Lösung

\int_{0}^{1} sin^{3} (x) d x

1 - cos (1) - \frac{1 - cos ^{3} ( 1 )}{3}

Dezimale

0.17894 \dots

Schritte zur Lösung

\int_{0}^{1} sin^{3} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{1} (1 - cos^{2} (x)) sin (x) d x

Wende U-Substitution an

= \int_{1}^{c o s (1)} - 1 + u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{c o s (1)}^{1} - 1 + u^{2} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int_{c o s (1)}^{1} 1 d u + \int_{c o s (1)}^{1} u^{2} d u)

\int_{c o s (1)}^{1} 1 d u = 1 - cos (1)

\int_{c o s (1)}^{1} u^{2} d u = \frac{1 - cos ^{3} ( 1 )}{3}

= - (- (1 - cos (1)) + \frac{1 - cos ^{3} ( 1 )}{3})

Vereinfache

= 1 - cos (1) - \frac{1 - cos ^{3} ( 1 )}{3}

\int_{1}^{9} 7 u d u

\int_{0}^{1} \frac{( y ^{2} ) ^{\frac{5}{2}}}{5} d y

\int_{- 6}^{- 3} (- x^{2} - 3 x) d x

\int_{- 4}^{4} (x^{2} - 4) d x

\int_{0}^{x^{6}} e^{- t} d t