integral of 8x^2cos(2x)

Lösung

\int 8 x^{2} cos (2 x) d x

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

Schritte zur Lösung

\int 8 x^{2} cos (2 x) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int x^{2} cos (2 x) d x

Wende die partielle Integration an

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int (x^{2} + 1)^{3} \cdot x d x

n = 1 \sum \infty (\frac{1}{n}) (\frac{1}{n})^{2}

y^{^{'}} = \frac{2 x y}{x ^{2} + 1}

t an g e n t f (x) = 6 + 1 - x, at x = - 8

\int \frac{x - 8}{( x + 4 ) ^{2} + 64} d x