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

The integral of x^2(a^2-x^2)^{1/2} is 1/8 a^4(arcsin(x/a)-1/4 sin(4arcsin(x/a)))+C

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

integral

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

\frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a}))) + C

Solution steps

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

Apply Trigonometric Substitution

: \int a^{4} sin^{2} (u) cos^{2} (u) d u

= \int a^{4} sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= a^{4} \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= a^{4} \cdot \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= a^{4} \frac{1}{8} \cdot \int 1 - cos (4 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^{4} \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= a^{4} \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

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

= a^{4} \frac{1}{8} (arcsin (\frac{1}{a} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{a} x)))

Simplify

a^{4} \frac{1}{8} (arcsin (\frac{1}{a} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{a} x))) : \frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a})))

= \frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a})))

Add a constant to the solution

= \frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a}))) + C

\frac{( 2 x - 1 )}{( 2 x + 3 )}

\frac{cos ( 2 x )}{cos ( x )}

\frac{( x + 8 ) ^{2}}{x ^{2} + 7 x - 8}

\frac{( 1 - cos ( x ) )}{sin ^{2} ( x )}

3 t^{2} + 5

What is the integral of x^2(a^2-x^2)^{1/2} ?
The integral of x^2(a^2-x^2)^{1/2} is 1/8 a^4(arcsin(x/a)-1/4 sin(4arcsin(x/a)))+C