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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-sqrt(1-x^2)

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

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

Solution steps

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

Take the constant out:

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

= - \int 1 - x^{2} d x

Apply Trigonometric Substitution

: \int cos^{2} (u) d u

= - \int cos^{2} (u) d u

Rewrite using trig identities

= - \int \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 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 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= - \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (x)

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

Add a constant to the solution

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

\int \frac{1}{2 x - 3 + 1} d x

\int \frac{sin ( 2 x )}{cos ( 2 x )} d x

\int (\frac{5 m - 2}{( m + 2 ) ^{2}}) d m

\int \frac{x ^{2}}{( 1 - x ) ^{100}} d x

\int x ln (\frac{1 + x}{1 - x}) d x

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