integral of (3-cos(x))^2

Lösung

\int (3 - cos (x))^{2} d x

- 6 sin (x) - \frac{19}{2} arcsin (cos (x)) + \frac{1}{4} sin (2 arcsin (cos (x))) + C

Schritte zur Lösung

\int (3 - cos (x))^{2} d x

Faktorisiere

3 - cos (x) : - (- 3 + cos (x))

= \int (- (- 3 + cos (x)))^{2} d x

Vereinfache

(- (- 3 + cos (x)))^{2} : (- 3 + cos (x))^{2}

= \int (- 3 + cos (x))^{2} d x

Wende U-Substitution an

= \int - \frac{( - 3 + u ) ^{2}}{1 - u ^{2}} d u

Entferne die Konstante:

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

= - \int \frac{( - 3 + u ) ^{2}}{1 - u ^{2}} d u

Multipliziere aus

\frac{( - 3 + u ) ^{2}}{1 - u ^{2}} : \frac{9}{1 - u ^{2}} - \frac{6 u}{1 - u ^{2}} + \frac{u ^{2}}{1 - u ^{2}}

= - \int \frac{9}{1 - u ^{2}} - \frac{6 u}{1 - u ^{2}} + \frac{u ^{2}}{1 - u ^{2}} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int \frac{9}{1 - u ^{2}} d u - \int \frac{6 u}{1 - u ^{2}} d u + \int \frac{u ^{2}}{1 - u ^{2}} d u)

\int \frac{9}{1 - u ^{2}} d u = 9 arcsin (u)

\int \frac{6 u}{1 - u ^{2}} d u = - 6 1 - u^{2}

\int \frac{u ^{2}}{1 - u ^{2}} d u = \frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

= - (9 arcsin (u) - (- 6 1 - u^{2}) + \frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))))

Setze in

u = cos (x)

ein

= - (9 arcsin (cos (x)) - (- 6 1 - cos^{2} (x)) + \frac{1}{2} (arcsin (cos (x)) - \frac{1}{2} sin (2 arcsin (cos (x)))))

Vereinfache

- (9 arcsin (cos (x)) - (- 6 1 - cos^{2} (x)) + \frac{1}{2} (arcsin (cos (x)) - \frac{1}{2} sin (2 arcsin (cos (x))))) : - 6 sin (x) - \frac{19}{2} arcsin (cos (x)) + \frac{1}{4} sin (2 arcsin (cos (x)))

= - 6 sin (x) - \frac{19}{2} arcsin (cos (x)) + \frac{1}{4} sin (2 arcsin (cos (x)))

Füge eine Konstante zur Lösung hinzu

= - 6 sin (x) - \frac{19}{2} arcsin (cos (x)) + \frac{1}{4} sin (2 arcsin (cos (x))) + C