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

Lösung

integral

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

\frac{1}{3} x^{3} cos^{2} (x) - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{4} x^{2} sin (2 x) + \frac{1}{4} x cos (2 x) + \frac{1}{2} x + \frac{1}{8} sin (2 x) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int x^{2} cos^{2} (x) d x = \frac{1}{3} x^{3} cos^{2} (x) - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{4} x^{2} sin (2 x) + \frac{1}{4} x cos (2 x) - \frac{1}{8} sin (2 x)

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

= \frac{1}{3} x^{3} cos^{2} (x) - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{4} x^{2} sin (2 x) + \frac{1}{4} x cos (2 x) - \frac{1}{8} sin (2 x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Vereinfache

\frac{1}{3} x^{3} cos^{2} (x) - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{4} x^{2} sin (2 x) + \frac{1}{4} x cos (2 x) - \frac{1}{8} sin (2 x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{1}{3} x^{3} cos^{2} (x) - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{4} x^{2} sin (2 x) + \frac{1}{4} x cos (2 x) + \frac{1}{2} x + \frac{1}{8} sin (2 x)

= \frac{1}{3} x^{3} cos^{2} (x) - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{4} x^{2} sin (2 x) + \frac{1}{4} x cos (2 x) + \frac{1}{2} x + \frac{1}{8} sin (2 x)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} x^{3} cos^{2} (x) - \frac{1}{6} x^{3} cos (2 x) + \frac{1}{4} x^{2} sin (2 x) + \frac{1}{4} x cos (2 x) + \frac{1}{2} x + \frac{1}{8} sin (2 x) + C

in t e g r a l \frac{r ^{2}}{1 + r ^{2}}

in t e g r a l \frac{1}{( x ^{2} + x + 1 ) ^{2}}

in t e g r a l \frac{x}{x ^{2}}

in t e g r a l \frac{1}{x ^{2} + 4 x + 9}

in t e g r a l \frac{1}{x ^{6}}