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

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

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integral of (a-x^2)^{1/2}

integral

(a - x^{2})^{\frac{1}{2}}

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= a \cdot \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

= a \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

= a \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)

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

sin^{3} (x) cos^{3} (x)

(\frac{( x - 1 )}{( x + 1 )})^{\frac{1}{2}}

sin^{2} (x) cos^{3} (x)

\frac{sin ( x )}{( 1 - sin ( x ) )}

\frac{1}{v ^{4}}

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