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

The integral of (1-2x^2)^3 is 1/35 x(-40x^6+84x^4-70x^2+35)+C

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

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

\frac{1}{35} x (- 40 x^{6} + 84 x^{4} - 70 x^{2} + 35) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Expand

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

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

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 1 d u - \int 3 u^{2} d u + \int 3 u^{4} d u - \int u^{6} d u)

\int 1 d u = u

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

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

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

= \frac{1}{2} (u - u^{3} + \frac{3 u ^{5}}{5} - \frac{u ^{7}}{7})

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} (2 x - (2 x)^{3} + \frac{3 ( 2 x ) ^{5}}{5} - \frac{( 2 x ) ^{7}}{7}) : \frac{1}{35} x (- 40 x^{6} + 84 x^{4} - 70 x^{2} + 35)

= \frac{1}{35} x (- 40 x^{6} + 84 x^{4} - 70 x^{2} + 35)

Add a constant to the solution

= \frac{1}{35} x (- 40 x^{6} + 84 x^{4} - 70 x^{2} + 35) + C

What is the integral of (1-2x^2)^3 ?
The integral of (1-2x^2)^3 is 1/35 x(-40x^6+84x^4-70x^2+35)+C