What is the integral of (1-x^2)^3 ?

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

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integral of (1-x^2)^3

integral

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

x - x^{3} + \frac{3 x ^{5}}{5} - \frac{x ^{7}}{7} + C

Solution steps

\int (1 - x^{2})^{3} d x

Expand

(1 - x^{2})^{3} : 1 - 3 x^{2} + 3 x^{4} - x^{6}

= \int 1 - 3 x^{2} + 3 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

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

\int 1 d x = x

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

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

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

= x - x^{3} + \frac{3 x ^{5}}{5} - \frac{x ^{7}}{7}

Add a constant to the solution

= x - x^{3} + \frac{3 x ^{5}}{5} - \frac{x ^{7}}{7} + C

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