What is the integral from 0 to x of sqrt(a^2-x^2) ?

The integral from 0 to x of sqrt(a^2-x^2) is a^2 1/2 (arcsin(x/a)+1/2 sin(2arcsin(x/a)))

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integral from 0 to x of sqrt(a^2-x^2)

\int_{0}^{x} a^{2} - x^{2} d x

a^{2} \frac{1}{2} (arcsin (\frac{x}{a}) + \frac{1}{2} sin (2 arcsin (\frac{x}{a})))

Solution steps

\int_{0}^{x} a^{2} - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{x}{a})} a^{2} cos^{2} (u) d u

= \int_{0}^{a r c s i n (\frac{x}{a})} a^{2} cos^{2} (u) d u

Take the constant out:

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

= a^{2} \cdot \int_{0}^{a r c s i n (\frac{x}{a})} cos^{2} (u) d u

Rewrite using trig identities

= a^{2} \cdot \int_{0}^{a r c s i n (\frac{x}{a})} \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

= a^{2} \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{x}{a})} 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

= a^{2} \frac{1}{2} (\int_{0}^{a r c s i n (\frac{x}{a})} 1 d u + \int_{0}^{a r c s i n (\frac{x}{a})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{x}{a})} 1 d u = arcsin (\frac{x}{a})

\int_{0}^{a r c s i n (\frac{x}{a})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{x}{a}))

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

\frac{d}{d x} (y^{2} - x^{3} + 8)

ln (x + 1)

X^{^{'}}

\frac{d y}{d x} = - \frac{1 + y ^{2}}{x y}

T \to 0 lim (a T + b)

What is the integral from 0 to x of sqrt(a^2-x^2) ?
The integral from 0 to x of sqrt(a^2-x^2) is a^2 1/2 (arcsin(x/a)+1/2 sin(2arcsin(x/a)))