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

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

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integral of sqrt(1-u^2)

\int 1 - u^{2} d u

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

Solution steps

\int 1 - u^{2} d u

Apply Trigonometric Substitution

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

= \int cos^{2} (v) d v

Rewrite using trig identities

= \int \frac{1 + cos ( 2 v )}{2} d v

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 v) d v

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 v + \int cos (2 v) d v)

\int 1 d v = v

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

= \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

v = arcsin (u)

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

Add a constant to the solution

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = \frac{82}{5}

\int 1 - 81 x^{2} d x

\int 7 cos^{2} (x) d x

e^{x} y \frac{d y}{d x} = e^{- y} e^{- 2 x - y}

\int \frac{x ^{2}}{x + 3} d x

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