integral of (sin^5(2x))/(cos^2(2x))

Lösung

\int \frac{sin ^{5} ( 2 x )}{cos ^{2} ( 2 x )} d x

- 160 (- \frac{1}{6} cos^{6} (x) + \frac{1}{4} cos^{8} (x) - \frac{1}{10} cos^{10} (x)) + \frac{1}{2} sin^{5} (2 x) tan (2 x) + C

Schritte zur Lösung

\int \frac{sin ^{5} ( 2 x )}{cos ^{2} ( 2 x )} d x

Wende die partielle Integration an

= - \int 5 sin^{5} (2 x) d x + \frac{1}{2} sin^{5} (2 x) tan (2 x)

\int 5 sin^{5} (2 x) d x = 160 (- \frac{1}{6} cos^{6} (x) + \frac{1}{4} cos^{8} (x) - \frac{1}{10} cos^{10} (x))

= - 160 (- \frac{1}{6} cos^{6} (x) + \frac{1}{4} cos^{8} (x) - \frac{1}{10} cos^{10} (x)) + \frac{1}{2} sin^{5} (2 x) tan (2 x)

Füge eine Konstante zur Lösung hinzu

= - 160 (- \frac{1}{6} cos^{6} (x) + \frac{1}{4} cos^{8} (x) - \frac{1}{10} cos^{10} (x)) + \frac{1}{2} sin^{5} (2 x) tan (2 x) + C

\frac{d}{d y} (e^{2 x y})

\int \frac{x ^{3}}{1 - x ^{8}} d x

d er i v a t i v e f (x) = 3 - 4 x

d er i v a t i v e y = 4 cos (x)

\frac{d y}{d x} = \frac{2 x ( y - 1 )}{x ^{2} + 1}