What is the integral of sqrt(2500-x^2) ?

The integral of sqrt(2500-x^2) is 1250(arcsin(1/50 x)+1/2 sin(2arcsin(1/50 x)))+C

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integral of sqrt(2500-x^2)

\int 2500 - x^{2} d x

1250 (arcsin (\frac{1}{50} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{50} x))) + C

Solution steps

\int 2500 - x^{2} d x

Apply Trigonometric Substitution

: \int 2500 cos^{2} (u) d u

= \int 2500 cos^{2} (u) d u

Take the constant out:

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

= 2500 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 2500 \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 2500 \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2500 \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)

= 2500 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{50} x)

= 2500 \cdot \frac{1}{2} (arcsin (\frac{1}{50} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{50} x)))

Simplify

2500 \cdot \frac{1}{2} (arcsin (\frac{1}{50} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{50} x))) : 1250 (arcsin (\frac{1}{50} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{50} x)))

= 1250 (arcsin (\frac{1}{50} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{50} x)))

Add a constant to the solution

= 1250 (arcsin (\frac{1}{50} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{50} x))) + C

\frac{d}{d x} ((2 x^{2} - 1)^{3} \cdot (1 - x^{2})^{- 3})

\frac{d}{d x} (2 + x^{2})

\frac{d}{d x} (\frac{( x ^{3} + 2 x ) ^{3}}{2 x - 4})

\frac{1}{cos ( x )}

\int 2 x + 4 d x

What is the integral of sqrt(2500-x^2) ?
The integral of sqrt(2500-x^2) is 1250(arcsin(1/50 x)+1/2 sin(2arcsin(1/50 x)))+C