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

Lösung

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

\frac{17}{8} x - \frac{1}{32} sin (4 x) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int 1 d x = x

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

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

\int sin^{2} (x) cos^{2} (x) d x = \frac{1}{8} (x - \frac{1}{4} sin (4 x))

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

Vereinfache

x + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : \frac{17}{8} x - \frac{1}{32} sin (4 x)

= \frac{17}{8} x - \frac{1}{32} sin (4 x)

Füge eine Konstante zur Lösung hinzu

= \frac{17}{8} x - \frac{1}{32} sin (4 x) + C

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