integral from 0 to pi of 5sin^2(tco)s^4t

Lösung

\int_{0}^{π} 5 sin^{2} (t co) s^{4} t d t

\frac{5 s ^{4} ( 2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co ) + 1 )}{8 c ^{2} o ^{2}}

Schritte zur Lösung

\int_{0}^{π} 5 sin^{2} (t co) s^{4} t d t

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 s^{4} \cdot \int_{0}^{π} sin^{2} (co t) t d t

Wende U-Substitution an

= 5 s^{4} \cdot \int_{0}^{co π} \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 s^{4} \frac{1}{c ^{2} o ^{2}} \cdot \int_{0}^{co π} u sin^{2} (u) d u

Wende die partielle Integration an

= 5 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u]_{0}^{π co}

\int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))

= 5 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))]_{0}^{π co}

Vereinfache

5 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))]_{0}^{π co} : \frac{5 s ^{4}}{c ^{2} o ^{2}} [\frac{2 u ^{2} - 2 u sin ( 2 u ) - cos ( 2 u )}{8}]_{0}^{π co}

= \frac{5 s ^{4}}{c ^{2} o ^{2}} [\frac{2 u ^{2} - 2 u sin ( 2 u ) - cos ( 2 u )}{8}]_{0}^{π co}

Berechne die Grenzen:

\frac{2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co )}{8} + \frac{1}{8}

= \frac{5 s ^{4}}{c ^{2} o ^{2}} (\frac{2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co )}{8} + \frac{1}{8})

Vereinfache

= \frac{5 s ^{4} ( 2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co ) + 1 )}{8 c ^{2} o ^{2}}

t an g e n t f (x) = (ln (x))^{4}, at x = 2

d er i v a t i v e y = \frac{x ^{3}}{5 - x ^{2}}

\frac{\partial}{\partial x} (ln (x^{2} + y^{4}))

a re a y = 2 x^{2} + 5 x + 6, (- 2, 0)

\frac{d}{d x} (\frac{1}{2} x^{2} - \frac{1}{4} x + \frac{1}{3})