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

解答

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

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

求解步骤

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

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

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

使用积分加法定则:

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

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

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

\int 1 d x = x

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

化简

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

解答补常数

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

\frac{d X}{d X}

l a pl a ce t r an s f or m 5 e^{- 3 (t - 2)}

y^{^{''}} - 2 y^{^{'}} + y = 2 t - 1 - (6 t e^{t} + 6 e^{t})

d er i v a t i v e 1 + T^{2}

\frac{d}{d x} (\frac{4 x - 3}{x ^{2} - 2 x})