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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

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

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

\int_{3}^{7} \frac{1}{( 7 + 2 x ) ^{3}} d x

\int_{2}^{4} \frac{1}{x ln ( x )} d x

\int_{- 3}^{3} v^{\frac{1}{3}} d v

\int_{2}^{4} \frac{x}{50} d x

\int_{0}^{t} sin (x) sin (t - x) d x

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