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

Lösung

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

\frac{3}{2} x - \frac{1}{8} sin (4 x) - cos (2 x) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

\int \frac{1}{2} d u = \frac{1}{2} u

\int sin (u) d u = - cos (u)

\int \frac{1}{2} sin^{2} (u) d u = \frac{1}{4} (u - \frac{1}{2} sin (2 u))

= \frac{1}{2} u - cos (u) + \frac{1}{4} (u - \frac{1}{2} sin (2 u))

Setze in

u = 2 x

ein

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

Vereinfache

\frac{1}{2} \cdot 2 x - cos (2 x) + \frac{1}{4} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) : \frac{3}{2} x - \frac{1}{8} sin (4 x) - cos (2 x)

= \frac{3}{2} x - \frac{1}{8} sin (4 x) - cos (2 x)

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} x - \frac{1}{8} sin (4 x) - cos (2 x) + C

\frac{d}{d t} (csc (t))

\int (sin (x))^{3} (cos (x))^{4} d x

(2 x^{3} + y) d x - x d y = 0

\frac{d}{d t} (1 + 2 cos (t))

\frac{d}{d x} (1 - 2 cos (x))