integral from 0 to 8pi of sqrt(5+t^2)

Lösung

\int_{0}^{8 π} 5 + t^{2} d t

4 π 64 π^{2} + 5 + \frac{5}{2} ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1)

Dezimale

324.86130 \dots

Schritte zur Lösung

\int_{0}^{8 π} 5 + t^{2} d t

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (\frac{8 π}{5})} 5 sec^{3} (u) d u

Entferne die Konstante:

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

= 5 \cdot \int_{0}^{a r c t a n (\frac{8 π}{5})} sec^{3} (u) d u

Wende integrale Reduktion an

= 5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{8 π}{5})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{8 π}{5})} sec (u) d u

\int_{0}^{a r c t a n (\frac{8 π}{5})} sec (u) d u = ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1)

= 5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{8 π}{5})} + \frac{1}{2} ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1)

Vereinfache

5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{8 π}{5})} + \frac{1}{2} ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1) : 5 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{8 π}{5})} + \frac{5}{2} ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1)

= 5 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{8 π}{5})} + \frac{5}{2} ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1)

Berechne die Grenzen:

\frac{4 π 64 π ^{2} + 5}{5}

= 5 \cdot \frac{4 π 64 π ^{2} + 5}{5} + \frac{5}{2} ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1)

Vereinfache

= 4 π 64 π^{2} + 5 + \frac{5}{2} ln (\frac{8 π}{5} + \frac{64 π ^{2}}{5} + 1)

d er i v a t i v e \frac{x}{e ^{x}}

l a pl a ce t r an s f or m f (t) = cos (t)

n = 0 \sum \infty \frac{1 0 ^{n}}{n !}

y^{^{'}} = - 5 y^{2}

f (x) = csc (x) cot (x)