integral of sin^4(x/2)

解答

\int sin^{4} (\frac{x}{2}) d x

2 (- \frac{1}{4} sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{3}{16} (x - sin (x))) + C

求解步骤

\int sin^{4} (\frac{x}{2}) d x

使用换元积分法

= \int sin^{4} (u) \cdot 2 d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int sin^{4} (u) d u

使用降次积分法

= 2 (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

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

= 2 (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

u = \frac{x}{2}

代回

= 2 (- \frac{cos ( \frac{x}{2} ) sin ^{3} ( \frac{x}{2} )}{4} + \frac{3}{4} \cdot \frac{1}{2} (\frac{x}{2} - \frac{1}{2} sin (2 \cdot \frac{x}{2})))

化简

2 (- \frac{cos ( \frac{x}{2} ) sin ^{3} ( \frac{x}{2} )}{4} + \frac{3}{4} \cdot \frac{1}{2} (\frac{x}{2} - \frac{1}{2} sin (2 \cdot \frac{x}{2}))) : 2 (- \frac{1}{4} sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{3}{16} (x - sin (x)))

= 2 (- \frac{1}{4} sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{3}{16} (x - sin (x)))

解答补常数

= 2 (- \frac{1}{4} sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{3}{16} (x - sin (x))) + C

x \to 4 lim (x^{2} + 2 x - 13)

x \to 1 + lim ((x - 1)^{\frac{1}{2}})

\frac{\partial}{\partial x} (\frac{( x + y )}{z})

\int - 4 e^{- x} d x

(22 + x^{2}) y^{^{'}} - 2 x y = 0