What is the integral of x^4sqrt(1-x^2) ?

The integral of x^4sqrt(1-x^2) is

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integral of x^4sqrt(1-x^2)

\int x^{4} 1 - x^{2} d x

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

Solution steps

\int x^{4} 1 - x^{2} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Expand

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

= \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

= \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)))

= - \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 (x)

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{7}

\int \frac{9 x - 27}{( x ^{2} + 2 x + 4 ) ^{2}} d x

\int_{0}^{5} \frac{x}{x + 10} d x

\int (x^{2} + 1) x d x

\frac{( d ^{2} y )}{( d x ^{2} )}