integral of 7sec^{-2}(xta)n^3x

Lösung

\int 7 sec^{- 2} (x t a) n^{3} x

\frac{7 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 7 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

= 7 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)

= 7 n^{3} \cdot \int x cos^{2} (a t x) d x

Wende U-Substitution an

= 7 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

= 7 n^{3} \frac{1}{a ^{2} t ^{2}} \cdot \int u cos^{2} (u) d u

Wende die partielle Integration an

= 7 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))

= 7 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

= 7 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

7 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{7 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{7 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{7 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

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

x \to 2 lim (x^{2} - 3 x + 6)

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

\int 10 x^{2} ln (x) d x

(y + x y) d x - 2 x d y = 0