integral of sin(1/(2x))

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用分布积分法

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

\int - \frac{cos ( u )}{u} d u = - Ci (u)

= - \frac{1}{2} (- \frac{sin ( u )}{u} - (- Ci (u)))

u = \frac{1}{2 x}

代回

= - \frac{1}{2} (- \frac{sin ( \frac{1}{2 x} )}{\frac{1}{2 x}} - (- Ci (\frac{1}{2 x})))

化简

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

= x sin (\frac{1}{2 x}) - \frac{1}{2} Ci (\frac{1}{2 x})

解答补常数

= x sin (\frac{1}{2 x}) - \frac{1}{2} Ci (\frac{1}{2 x}) + C

t an g e n t y = 16 x

\int \frac{3 x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

\int 6 t^{2} d t

d er i v a t i v e x^{3} e^{- 2 x}

\int (\frac{1}{x ^{2}} - x^{2} - \frac{1}{3}) d x