integral of 35sec^{-2}(xta)n^3x

Lösung

\int 35 sec^{- 2} (x t a) n^{3} x d x

\frac{35 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x)) + C

Schritte zur Lösung

\int 35 sec^{- 2} (x t a) n^{3} x d x

Entferne die Konstante:

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

= 35 n^{3} \cdot \int sec^{- 2} (a t x) x d x

Vereinfache

sec^{- 2} (a t x) x : x cos^{2} (a t x)

= 35 n^{3} \cdot \int x cos^{2} (a t x) d x

Wende U-Substitution an

= 35 n^{3} \cdot \int \frac{u cos ^{2} ( u )}{a ^{2} t ^{2}} d u

Entferne die Konstante:

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

= 35 n^{3} \frac{1}{a ^{2} t ^{2}} \cdot \int u cos^{2} (u) d u

Wende die partielle Integration an

= 35 n^{3} \frac{1}{a ^{2} t ^{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)

\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))

= 35 n^{3} \frac{1}{a ^{2} t ^{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)))

Setze in

u = a t x

ein

= 35 n^{3} \frac{1}{a ^{2} t ^{2}} (\frac{1}{2} a t x (a t x + \frac{1}{2} sin (2 a t x)) - \frac{1}{2} (\frac{( a t x ) ^{2}}{2} - \frac{1}{4} cos (2 a t x)))

Vereinfache

35 n^{3} \frac{1}{a ^{2} t ^{2}} (\frac{1}{2} a t x (a t x + \frac{1}{2} sin (2 a t x)) - \frac{1}{2} (\frac{( a t x ) ^{2}}{2} - \frac{1}{4} cos (2 a t x))) : \frac{35 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x))

= \frac{35 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x))

Füge eine Konstante zur Lösung hinzu

= \frac{35 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x)) + C

x \to 1 lim (\frac{1 - x}{x ^{2} - 1})

(- \frac{1}{x})^{^{'}}

y^{^{''}} + 2 y^{^{'}} = 2 x + 9 - e^{- 2 x}

in v erse l a pl a ce \frac{6 s + 4}{s ^{2} + 4 s + 20}

y^{^{'}} + 5 y = e^{- 3 x}