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

The integral of ((a^2-x^2)^3)/2 is 1/2 (a^6x-a^4x^3+(3a^2x^5)/5-(x^7)/7)+C

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

integral

\frac{( a ^{2} - x ^{2} ) ^{3}}{2}

\frac{1}{2} (a^{6} x - a^{4} x^{3} + \frac{3 a ^{2} x ^{5}}{5} - \frac{x ^{7}}{7}) + C

Solution steps

\int \frac{( a ^{2} - x ^{2} ) ^{3}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int (a^{2} - x^{2})^{3} d x

Expand

(a^{2} - x^{2})^{3} : a^{6} - 3 a^{4} x^{2} + 3 a^{2} x^{4} - x^{6}

= \frac{1}{2} \cdot \int a^{6} - 3 a^{4} x^{2} + 3 a^{2} x^{4} - x^{6} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int a^{6} d x - \int 3 a^{4} x^{2} d x + \int 3 a^{2} x^{4} d x - \int x^{6} d x)

\int a^{6} d x = a^{6} x

\int 3 a^{4} x^{2} d x = a^{4} x^{3}

\int 3 a^{2} x^{4} d x = \frac{3 a ^{2} x ^{5}}{5}

\int x^{6} d x = \frac{x ^{7}}{7}

= \frac{1}{2} (a^{6} x - a^{4} x^{3} + \frac{3 a ^{2} x ^{5}}{5} - \frac{x ^{7}}{7})

Add a constant to the solution

= \frac{1}{2} (a^{6} x - a^{4} x^{3} + \frac{3 a ^{2} x ^{5}}{5} - \frac{x ^{7}}{7}) + C

\frac{sinh ( x )}{( 1 + cosh ^{2} ( x ) )}

\frac{1}{( 1 + t ) ^{4}}

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

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

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

What is the integral of ((a^2-x^2)^3)/2 ?
The integral of ((a^2-x^2)^3)/2 is 1/2 (a^6x-a^4x^3+(3a^2x^5)/5-(x^7)/7)+C