integral from 0 to sin(2x) of arcsin(t)

Lösung

\int_{0}^{s i n (2 x)} arcsin (t) d t

cos (2 x) + sin (2 x) arcsin (sin (2 x)) - 1

Schritte zur Lösung

\int_{0}^{s i n (2 x)} arcsin (t) d t

Wende die partielle Integration an

= [t arcsin (t) - \int \frac{t}{1 - t ^{2}} d t]_{0}^{s i n (2 x)}

\int \frac{t}{1 - t ^{2}} d t = - 1 - t^{2}

= [t arcsin (t) - (- 1 - t^{2})]_{0}^{s i n (2 x)}

Vereinfache

= [t arcsin (t) + 1 - t^{2}]_{0}^{s i n (2 x)}

Berechne die Grenzen:

cos (2 x) + sin (2 x) arcsin (sin (2 x)) - 1

= cos (2 x) + sin (2 x) arcsin (sin (2 x)) - 1

\int_{2}^{3} x^{2} + x + 1 d x

\int_{x}^{\infty} e^{- x} d x

\int_{1}^{3} x^{2} - 6 x + 5 d x

\int_{2}^{3} x^{2} + x - 2 d x

\int_{- 2}^{2} 4^{\frac{x}{2}} d x