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受欢迎的微积分 >

integral of (u^2)/(2sqrt(1-u^2))

解答

\int \frac{u ^{2}}{2 1 - u ^{2}} d u

解答

\frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) + C

求解步骤

\int \frac{u ^{2}}{2 1 - 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{u ^{2}}{1 - u ^{2}} d u

使用三角换元法

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

使用三角恒等式改写

= \frac{1}{2} \cdot \int \frac{1 - cos ( 2 v )}{2} d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 - cos (2 v) d v

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

= \frac{1}{2} \cdot \frac{1}{2} (\int 1 d v - \int cos (2 v) d v)

\int 1 d v = v

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

= \frac{1}{2} \cdot \frac{1}{2} (v - \frac{1}{2} sin (2 v))

v = arcsin (u)

代回

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

化简

\frac{1}{2} \cdot \frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) : \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

= \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

解答补常数

= \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) + C

作图

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