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integral of (128)/(w^2sqrt(64-w^2))

解答

\int \frac{128}{w ^{2} 64 - w ^{2}} d w

解答

- \frac{2 - w ^{2} + 64}{w} + C

求解步骤

\int \frac{128}{w ^{2} 64 - w ^{2}} d w

提出常数:

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

= 128 \cdot \int \frac{1}{w ^{2} 64 - w ^{2}} d w

使用三角换元法

= 128 \cdot \int \frac{cot ( u )}{64 cos ( u ) sin ( u )} d u

提出常数:

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

= 128 \cdot \frac{1}{64} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

化简

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= 128 \cdot \frac{1}{64} \cdot \int csc^{2} (u) d u

使用积分常见定则:

\int csc^{2} (u) d u = - cot (u)

= 128 \cdot \frac{1}{64} (- cot (u))

u = arcsin (\frac{1}{8} w)

代回

= 128 \cdot \frac{1}{64} (- cot (arcsin (\frac{1}{8} w)))

化简

128 \cdot \frac{1}{64} (- cot (arcsin (\frac{1}{8} w))) : - \frac{2 - w ^{2} + 64}{w}

= - \frac{2 - w ^{2} + 64}{w}

解答补常数

= - \frac{2 - w ^{2} + 64}{w} + C

作图

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