What is the integral of x^3sqrt(x^2-a^2) ?

The integral of x^3sqrt(x^2-a^2) is (a^2(x^2-a^2)^{3/2})/3+1/5 (x^2-a^2)^{5/2}+C

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integral of x^3sqrt(x^2-a^2)

\int x^{3} x^{2} - a^{2} d x

\frac{a ^{2} ( x ^{2} - a ^{2} ) ^{\frac{3}{2}}}{3} + \frac{1}{5} (x^{2} - a^{2})^{\frac{5}{2}} + C

Solution steps

\int x^{3} x^{2} - a^{2} d x

Apply Trigonometric Substitution

= \int a^{5} sec^{4} (u) tan^{2} (u) d u

Take the constant out:

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

= a^{5} \cdot \int sec^{4} (u) tan^{2} (u) d u

Rewrite using trig identities

= a^{5} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

Apply u-substitution

= a^{5} \cdot \int v^{2} (1 + v^{2}) d v

Expand

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= a^{5} \cdot \int v^{2} + v^{4} 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^{5} (\int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= a^{5} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= a^{5} (\frac{tan ^{3} ( arcsec ( \frac{1}{a} x ) )}{3} + \frac{tan ^{5} ( arcsec ( \frac{1}{a} x ) )}{5})

Simplify

a^{5} (\frac{tan ^{3} ( arcsec ( \frac{1}{a} x ) )}{3} + \frac{tan ^{5} ( arcsec ( \frac{1}{a} x ) )}{5}) : \frac{a ^{2} ( x ^{2} - a ^{2} ) ^{\frac{3}{2}}}{3} + \frac{1}{5} (x^{2} - a^{2})^{\frac{5}{2}}

= \frac{a ^{2} ( x ^{2} - a ^{2} ) ^{\frac{3}{2}}}{3} + \frac{1}{5} (x^{2} - a^{2})^{\frac{5}{2}}

Add a constant to the solution

= \frac{a ^{2} ( x ^{2} - a ^{2} ) ^{\frac{3}{2}}}{3} + \frac{1}{5} (x^{2} - a^{2})^{\frac{5}{2}} + C

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What is the integral of x^3sqrt(x^2-a^2) ?
The integral of x^3sqrt(x^2-a^2) is (a^2(x^2-a^2)^{3/2})/3+1/5 (x^2-a^2)^{5/2}+C