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

Lösung

\int_{0}^{8 π} x^{2} + 5 d x

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} x^{2} + 5 d x

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)

\int \frac{5}{( x ^{2} + 4 )} d x

\frac{d}{d x} (\frac{1 - cos ( 2 x )}{1 + cos ( 2 x )})

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

\frac{d}{d x} (4 e^{- 6 x^{2}})

\int \frac{5 ( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )} d x