What is the integral of (1-x^2)^{0.5} ?

The integral of (1-x^2)^{0.5} is 1/2 (arcsin(x)+1/2 sin(2arcsin(x)))+C

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integral of (1-x^2)^{0.5}

\int (1 - x^{2})^{0.5} d x

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

Solution steps

\int (1 - x^{2})^{0.5} d x

Simplify

x^{2} : xx

= \int (1 - xx)^{0.5} d x

xx = x^{2}

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

\frac{2}{4 - x}, at x = 1

\int sin^{2} (4 x) cos^{2} (x) 4 x d x

\frac{d y}{d x} = 5 x e^{y}, y (0) = 0

\int \frac{x ^{3}}{3 - x ^{4}} d x

\frac{d}{d x} (12 x tanh (x))

What is the integral of (1-x^2)^{0.5} ?
The integral of (1-x^2)^{0.5} is 1/2 (arcsin(x)+1/2 sin(2arcsin(x)))+C