integral of (sin(2x))/(1+sin(2x))

解

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

x + \frac{1}{tan ( x ) + 1} + C

解答ステップ

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

長除法

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

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

総和規則を適用する:

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

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

\int 1 d x = x

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

= x - (- \frac{1}{tan ( x ) + 1})

簡素化

= x + \frac{1}{tan ( x ) + 1}

定数を解答に追加する

= x + \frac{1}{tan ( x ) + 1} + C