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integral of 5sqrt(1-25x^2)

解答

\int 5 1 - 25 x^{2} d x

解答

\frac{1}{2} (arcsin (5 x) + \frac{1}{2} sin (2 arcsin (5 x))) + C

求解步骤

\int 5 1 - 25 x^{2} d x

提出常数:

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

= 5 \cdot \int 1 - 25 x^{2} d x

使用三角换元法

= 5 \cdot \int \frac{1}{5} cos^{2} (u) d u

提出常数:

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

= 5 \cdot \frac{1}{5} \cdot \int cos^{2} (u) d u

使用三角恒等式改写

= 5 \cdot \frac{1}{5} \cdot \int \frac{1 + cos ( 2 u )}{2} d u

提出常数:

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

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

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

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

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

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

u = arcsin (5 x)

代回

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} (arcsin (5 x) + \frac{1}{2} sin (2 arcsin (5 x)))

化简

5 \cdot \frac{1}{5} \cdot \frac{1}{2} (arcsin (5 x) + \frac{1}{2} sin (2 arcsin (5 x))) : \frac{1}{2} (arcsin (5 x) + \frac{1}{2} sin (2 arcsin (5 x)))

= \frac{1}{2} (arcsin (5 x) + \frac{1}{2} sin (2 arcsin (5 x)))

解答补常数

= \frac{1}{2} (arcsin (5 x) + \frac{1}{2} sin (2 arcsin (5 x))) + C

作图

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