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

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

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integral of sqrt(a^2-u^2)

\int a^{2} - u^{2} d u

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

Solution steps

\int a^{2} - u^{2} d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= a^{2} \cdot \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

= a^{2} \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

= a^{2} \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)

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

Substitute back

v = arcsin (\frac{1}{a} u)

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

Simplify

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

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

Add a constant to the solution

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

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