integral of sin^2(x)+3

Lösung

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

\frac{7}{2} x - \frac{1}{4} sin (2 x) + C

Schritte zur Lösung

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

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

\int 3 d x = 3 x

= \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + 3 x

Vereinfache

\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + 3 x : \frac{7}{2} x - \frac{1}{4} sin (2 x)

= \frac{7}{2} x - \frac{1}{4} sin (2 x)

Füge eine Konstante zur Lösung hinzu

= \frac{7}{2} x - \frac{1}{4} sin (2 x) + C

\int \frac{1}{x ^{4} 4 x ^{2} - 9} d x

\int - \frac{2}{81 + 25 t ^{2}} d t

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

\int (π x - x^{2}) sin (n x) d x

\int \frac{e ^{x}}{e ^{2 x} - 4 e ^{x} + 6} d x