What is the integral from 0 to y of sqrt(1-y^2) ?

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

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integral from 0 to y of sqrt(1-y^2)

\int_{0}^{y} 1 - y^{2} d y

\frac{1}{2} (arcsin (y) + \frac{1}{2} sin (2 arcsin (y)))

Solution steps

\int_{0}^{y} 1 - y^{2} d y

Apply Trigonometric Substitution

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

Rewrite using trig identities

= \int_{0}^{a r c s i n (y)} \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

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

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

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

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

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

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