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

Lösung

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

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

Schritte zur Lösung

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

\int 1 d x = x

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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

2 x y \frac{d y}{d x} = 5

x \to 2 lim (3 x + 4)

f (t) = sin (t)

\frac{d}{d x} (f (x) (sin (x)))