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

The integral of (1-x^2)^4 is x-(4x^3)/3+(6x^5)/5-(4x^7)/7+(x^9)/9+C

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integral of (1-x^2)^4

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

x - \frac{4 x ^{3}}{3} + \frac{6 x ^{5}}{5} - \frac{4 x ^{7}}{7} + \frac{x ^{9}}{9} + C

Solution steps

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

Expand

(1 - x^{2})^{4} : 1 - 4 x^{2} + 6 x^{4} - 4 x^{6} + x^{8}

= \int 1 - 4 x^{2} + 6 x^{4} - 4 x^{6} + x^{8} 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 4 x^{2} d x + \int 6 x^{4} d x - \int 4 x^{6} d x + \int x^{8} d x

\int 1 d x = x

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

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int x^{8} d x = \frac{x ^{9}}{9}

= x - \frac{4 x ^{3}}{3} + \frac{6 x ^{5}}{5} - \frac{4 x ^{7}}{7} + \frac{x ^{9}}{9}

Add a constant to the solution

= x - \frac{4 x ^{3}}{3} + \frac{6 x ^{5}}{5} - \frac{4 x ^{7}}{7} + \frac{x ^{9}}{9} + C

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What is the integral of (1-x^2)^4 ?
The integral of (1-x^2)^4 is x-(4x^3)/3+(6x^5)/5-(4x^7)/7+(x^9)/9+C