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

The integral of 1/3 (1-x^2)^3 is 1/3 x-1/3 x^3+(x^5)/5-(x^7)/(21)+C

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integral of 1/3 (1-x^2)^3

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

\frac{1}{3} x - \frac{1}{3} x^{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{21} + C

Solution steps

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

Take the constant out:

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

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

Expand

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

= \frac{1}{3} \cdot \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

= \frac{1}{3} (\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}

= \frac{1}{3} (x - x^{3} + \frac{3 x ^{5}}{5} - \frac{x ^{7}}{7})

Simplify

\frac{1}{3} (x - x^{3} + \frac{3 x ^{5}}{5} - \frac{x ^{7}}{7}) : \frac{1}{3} x - \frac{1}{3} x^{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{21}

= \frac{1}{3} x - \frac{1}{3} x^{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{21}

Add a constant to the solution

= \frac{1}{3} x - \frac{1}{3} x^{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{21} + C

cos (x) \frac{d y}{d x} + (sin (x)) y = 1

\frac{\partial}{\partial y} (e^{x^{2} + y^{2}} sin (x y))

\int \frac{5 + 5 x + 5}{x + 5} d x

(x^{2} + 1) \frac{d y}{d x} + 5 x (y - 1) = 0

2 y^{^{''}} - 11 y^{^{'}} - 6 y = 0

What is the integral of 1/3 (1-x^2)^3 ?
The integral of 1/3 (1-x^2)^3 is 1/3 x-1/3 x^3+(x^5)/5-(x^7)/(21)+C