What is the integral of x^4(a^2-x^2)^{1/2} ?

The integral of x^4(a^2-x^2)^{1/2} is

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integral of x^4(a^2-x^2)^{1/2}

integral

x^{4} (a^{2} - x^{2})^{\frac{1}{2}}

- \frac{a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{4} + \frac{3}{8} a^{6} arcsin (\frac{x}{a}) - \frac{3}{16} a^{6} sin (2 arcsin (\frac{x}{a})) + \frac{1}{6} x^{5} (a^{2} - x^{2})^{\frac{1}{2}} + \frac{5 a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{24} - \frac{5 a ^{6}}{32} (2 arcsin (\frac{x}{a}) - sin (2 arcsin (\frac{x}{a}))) + C

Solution steps

\int x^{4} (a^{2} - x^{2})^{\frac{1}{2}} d x

Apply Trigonometric Substitution

: \int a^{6} sin^{4} (u) cos^{2} (u) d u

= \int a^{6} sin^{4} (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^{6} \cdot \int sin^{4} (u) cos^{2} (u) d u

Rewrite using trig identities

= a^{6} \cdot \int sin^{4} (u) (1 - sin^{2} (u)) d u

Expand

sin^{4} (u) (1 - sin^{2} (u)) : sin^{4} (u) - sin^{6} (u)

= a^{6} \cdot \int sin^{4} (u) - sin^{6} (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^{6} (\int sin^{4} (u) d u - \int sin^{6} (u) d u)

\int sin^{4} (u) d u = - \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u))

\int sin^{6} (u) d u = - \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)))

= a^{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)) - (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)))))

Substitute back

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

= a^{6} (- \frac{1}{4} sin^{3} (arcsin (\frac{1}{a} x)) cos (arcsin (\frac{1}{a} x)) + \frac{3}{8} (arcsin (\frac{1}{a} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) - (- \frac{cos ( arcsin ( \frac{1}{a} x ) ) sin ^{5} ( arcsin ( \frac{1}{a} x ) )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (arcsin (\frac{1}{a} x)) cos (arcsin (\frac{1}{a} x)) + \frac{3}{8} (arcsin (\frac{1}{a} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))))))

Simplify

a^{6} (- \frac{1}{4} sin^{3} (arcsin (\frac{1}{a} x)) cos (arcsin (\frac{1}{a} x)) + \frac{3}{8} (arcsin (\frac{1}{a} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) - (- \frac{cos ( arcsin ( \frac{1}{a} x ) ) sin ^{5} ( arcsin ( \frac{1}{a} x ) )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (arcsin (\frac{1}{a} x)) cos (arcsin (\frac{1}{a} x)) + \frac{3}{8} (arcsin (\frac{1}{a} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{a} x)))))) : - \frac{a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{4} + \frac{3}{8} a^{6} arcsin (\frac{x}{a}) - \frac{3}{16} a^{6} sin (2 arcsin (\frac{x}{a})) + \frac{1}{6} x^{5} (a^{2} - x^{2})^{\frac{1}{2}} + \frac{5 a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{24} - \frac{5 a ^{6}}{32} (2 arcsin (\frac{x}{a}) - sin (2 arcsin (\frac{x}{a})))

= - \frac{a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{4} + \frac{3}{8} a^{6} arcsin (\frac{x}{a}) - \frac{3}{16} a^{6} sin (2 arcsin (\frac{x}{a})) + \frac{1}{6} x^{5} (a^{2} - x^{2})^{\frac{1}{2}} + \frac{5 a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{24} - \frac{5 a ^{6}}{32} (2 arcsin (\frac{x}{a}) - sin (2 arcsin (\frac{x}{a})))

Add a constant to the solution

= - \frac{a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{4} + \frac{3}{8} a^{6} arcsin (\frac{x}{a}) - \frac{3}{16} a^{6} sin (2 arcsin (\frac{x}{a})) + \frac{1}{6} x^{5} (a^{2} - x^{2})^{\frac{1}{2}} + \frac{5 a ^{2} x ^{3} ( - x ^{2} + a ^{2} ) ^{\frac{1}{2}}}{24} - \frac{5 a ^{6}}{32} (2 arcsin (\frac{x}{a}) - sin (2 arcsin (\frac{x}{a}))) + C

tan (x) + x^{6} + e^{x}

\frac{1}{( 3 x ^{2} )}

\frac{( x ^{5} - 1 )}{x - 1}

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

3 x^{7}

What is the integral of x^4(a^2-x^2)^{1/2} ?
The integral of x^4(a^2-x^2)^{1/2} is