integral of (1+cos(x))^2

Lösung

\int (1 + cos (x))^{2} d x

\frac{3}{2} x + \frac{1}{4} sin (2 x) + 2 sin (x) + C

Schritte zur Lösung

\int (1 + cos (x))^{2} d x

Multipliziere aus

(1 + cos (x))^{2} : 1 + 2 cos (x) + cos^{2} (x)

= \int 1 + 2 cos (x) + cos^{2} (x) d x

Wende die Summenregel an:

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

= \int 1 d x + \int 2 cos (x) d x + \int cos^{2} (x) d x

\int 1 d x = x

\int 2 cos (x) d x = 2 sin (x)

\int cos^{2} (x) d x = \frac{1}{2} (x + \frac{1}{2} sin (2 x))

= x + 2 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Vereinfache

x + 2 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{3}{2} x + \frac{1}{4} sin (2 x) + 2 sin (x)

= \frac{3}{2} x + \frac{1}{4} sin (2 x) + 2 sin (x)

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} x + \frac{1}{4} sin (2 x) + 2 sin (x) + C

\frac{\partial}{\partial x} (cos (2 x) sin (3 y))

t an g e n t f (x) = 2 x^{3} + 3 x^{2} - 12 x + 1

\frac{d ^{2} y}{d x ^{2}} = 4 x cos (x^{2})

3 y^{^{'}} + y = (1 - 2 x) y^{4}

y = \frac{- 1}{3 x ^{2}}