integral of (sin(θ))/(1+sin(θ))

Lösung

\int \frac{sin ( θ )}{1 + sin ( θ )} d θ

\frac{2}{tan ( \frac{θ}{2} ) + 1} + θ + C

Schritte zur Lösung

\int \frac{sin ( θ )}{1 + sin ( θ )} d θ

\frac{sin ( θ )}{1 + sin ( θ )} = - \frac{1}{1 + sin ( θ )} + 1

= \int - \frac{1}{1 + sin ( θ )} + 1 d θ

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{1 + sin ( θ )} d θ + \int 1 d θ

\int \frac{1}{1 + sin ( θ )} d θ = - \frac{2}{tan ( \frac{θ}{2} ) + 1}

\int 1 d θ = θ

= - (- \frac{2}{tan ( \frac{θ}{2} ) + 1}) + θ

Vereinfache

= \frac{2}{tan ( \frac{θ}{2} ) + 1} + θ

Füge eine Konstante zur Lösung hinzu

= \frac{2}{tan ( \frac{θ}{2} ) + 1} + θ + C

\int \frac{- 2}{x ^{2} + 1} d x

\int 2 x + 0 d x

\int sin (ln (x)) d x

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