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受欢迎的微积分 >

integral of (a^2)/(sqrt(4a-a^2))

解答

\int \frac{a ^{2}}{4 a - a ^{2}} d a

解答

6 arcsin (\frac{1}{2} a - 1) - sin (2 arcsin (\frac{1}{2} a - 1)) - 4 - a^{2} + 4 a + C

求解步骤

\int \frac{a ^{2}}{4 a - a ^{2}} d a

对

4 a - a^{2}

配方:

- (a - 2)^{2} + 4

= \int \frac{a ^{2}}{- ( a - 2 ) ^{2} + 4} d a

使用换元积分法

= \int \frac{( u + 2 ) ^{2}}{- u ^{2} + 4} d u

使用三角换元法

= \int (2 sin (v) + 2)^{2} d v

乘开

(2 sin (v) + 2)^{2} : 4 sin^{2} (v) + 8 sin (v) + 4

= \int 4 sin^{2} (v) + 8 sin (v) + 4 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

= \int 4 sin^{2} (v) d v + \int 8 sin (v) d v + \int 4 d v

\int 4 sin^{2} (v) d v = 2 v - sin (2 v)

\int 8 sin (v) d v = - 8 cos (v)

\int 4 d v = 4 v

= 2 v - sin (2 v) - 8 cos (v) + 4 v

代回

= 2 arcsin (\frac{1}{2} (a - 2)) - sin (2 arcsin (\frac{1}{2} (a - 2))) - 8 cos (arcsin (\frac{1}{2} (a - 2))) + 4 arcsin (\frac{1}{2} (a - 2))

化简

2 arcsin (\frac{1}{2} (a - 2)) - sin (2 arcsin (\frac{1}{2} (a - 2))) - 8 cos (arcsin (\frac{1}{2} (a - 2))) + 4 arcsin (\frac{1}{2} (a - 2)) : 6 arcsin (\frac{1}{2} a - 1) - sin (2 arcsin (\frac{1}{2} a - 1)) - 4 - a^{2} + 4 a

= 6 arcsin (\frac{1}{2} a - 1) - sin (2 arcsin (\frac{1}{2} a - 1)) - 4 - a^{2} + 4 a

解答补常数

= 6 arcsin (\frac{1}{2} a - 1) - sin (2 arcsin (\frac{1}{2} a - 1)) - 4 - a^{2} + 4 a + C

作图

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