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

Lösung

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

\frac{1}{2} x^{2} cos (2 x + 1) - \frac{1}{2} x sin (2 x + 1) - \frac{5}{4} cos (2 x + 1) + C

Schritte zur Lösung

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

Wende die partielle Integration an

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

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

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

Vereinfache

- cos (2 x + 1) + \frac{1}{2} x^{2} cos (2 x + 1) - (\frac{1}{2} x sin (2 x + 1) + \frac{1}{4} cos (2 x + 1)) : \frac{1}{2} x^{2} cos (2 x + 1) - \frac{1}{2} x sin (2 x + 1) - \frac{5}{4} cos (2 x + 1)

= \frac{1}{2} x^{2} cos (2 x + 1) - \frac{1}{2} x sin (2 x + 1) - \frac{5}{4} cos (2 x + 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} x^{2} cos (2 x + 1) - \frac{1}{2} x sin (2 x + 1) - \frac{5}{4} cos (2 x + 1) + C

\int e^{3 x} (e^{3 x} + 4)^{- 2} d x

\int \frac{7 x + 1}{6 x ^{2} + x + 1} d x

\int (4 x^{2} + 1)^{\frac{3}{2}} d x

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

\int \frac{x}{x ^{2} - 2 a x + 2 a ^{2}} d x