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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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_{0}^{2} 2 x + x^{2} - x^{3} d x

\int_{0}^{5} ∣ 1 - x ∣ d x

\int_{0}^{2} (t - 1) e^{(t - 1)^{2}} d t

\int_{- 4}^{4} - x^{2} - 3 x d x

\int_{0}^{2} (3 t - 1)^{56} d t

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