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受欢迎的微积分 >

integral of u 1/(sqrt(4-9u^2))

解答

\int u \frac{1}{4 - 9 u ^{2}} d u

解答

- \frac{1}{9} 4 - 9 u^{2} + C

求解步骤

\int u \frac{1}{4 - 9 u ^{2}} d u

化简

u \frac{1}{4 - 9 u ^{2}} : \frac{u}{4 - 9 u ^{2}}

= \int \frac{u}{4 - 9 u ^{2}} d u

使用换元积分法

= \int - \frac{1}{18 v} d v

提出常数:

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

= - \frac{1}{18} \cdot \int \frac{1}{v} d v

使用幂法则

= - \frac{1}{18} \cdot 2 v^{\frac{1}{2}}

v = 4 - 9 u^{2}

代回

= - \frac{1}{18} \cdot 2 (4 - 9 u^{2})^{\frac{1}{2}}

化简

- \frac{1}{18} \cdot 2 (4 - 9 u^{2})^{\frac{1}{2}} : - \frac{1}{9} 4 - 9 u^{2}

= - \frac{1}{9} 4 - 9 u^{2}

解答补常数

= - \frac{1}{9} 4 - 9 u^{2} + C

作图

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\int cos^{3} (y) sin^{3} (y) d y

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