integral of sin^2(wt)

解答

\int sin^{2} (wt) d t

\frac{1}{2} t - \frac{1}{4 w} sin (2 wt) + C

求解步骤

\int sin^{2} (wt) d t

使用三角恒等式改写

= \int 1 - cos^{2} (wt) d t

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 1 d t - \int cos^{2} (wt) d t

\int 1 d t = t

\int cos^{2} (wt) d t = \frac{1}{2} (t + \frac{1}{2 w} sin (2 wt))

= t - \frac{1}{2} (t + \frac{1}{2 w} sin (2 wt))

化简

t - \frac{1}{2} (t + \frac{1}{2 w} sin (2 wt)) : \frac{1}{2} t - \frac{1}{4 w} sin (2 wt)

= \frac{1}{2} t - \frac{1}{4 w} sin (2 wt)

解答补常数

= \frac{1}{2} t - \frac{1}{4 w} sin (2 wt) + C

ma c l a u r in arctan (4 x)

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

\frac{\partial}{\partial x} (l - c)

\int x (15 x + 4) d x

d er i v a t i v e \frac{( 6 x ^{2} + 6 x + 4 )}{x}