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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

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

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

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

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

\int_{- \infty}^{1} 3 e^{\frac{x}{2}} d x

\int_{0}^{y} x e^{y^{2}} d x

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

\int_{0}^{4} ∣ 1 - 2 x ∣ d x

\int_{- 1}^{0} \frac{3 t}{( 5 + t ^{2} ) ^{4}} d t

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