What is the integral of x^3(1-4x^2)^{3/2} ?

The integral of x^3(1-4x^2)^{3/2} is -1/280 (10x^2+1)(-4x^2+1)^{5/2}+C

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integral of x^3(1-4x^2)^{3/2}

\int x^{3} (1 - 4 x^{2})^{\frac{3}{2}} d x

- \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}} + C

Solution steps

\int x^{3} (1 - 4 x^{2})^{\frac{3}{2}} d x

Apply u-substitution

= \int \frac{u ^{\frac{3}{2}} ( u - 1 )}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int u^{\frac{3}{2}} (u - 1) d u

Apply Integration By Parts

= \frac{1}{32} (\frac{2}{5} u^{\frac{5}{2}} (u - 1) - \int \frac{2}{5} u^{\frac{5}{2}} d u)

\int \frac{2}{5} u^{\frac{5}{2}} d u = \frac{4}{35} u^{\frac{7}{2}}

= \frac{1}{32} (\frac{2}{5} u^{\frac{5}{2}} (u - 1) - \frac{4}{35} u^{\frac{7}{2}})

Substitute back

u = 1 - 4 x^{2}

= \frac{1}{32} (\frac{2}{5} (1 - 4 x^{2})^{\frac{5}{2}} (1 - 4 x^{2} - 1) - \frac{4}{35} (1 - 4 x^{2})^{\frac{7}{2}})

Simplify

\frac{1}{32} (\frac{2}{5} (1 - 4 x^{2})^{\frac{5}{2}} (1 - 4 x^{2} - 1) - \frac{4}{35} (1 - 4 x^{2})^{\frac{7}{2}}) : - \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}}

= - \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}}

Add a constant to the solution

= - \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}} + C

What is the integral of x^3(1-4x^2)^{3/2} ?
The integral of x^3(1-4x^2)^{3/2} is -1/280 (10x^2+1)(-4x^2+1)^{5/2}+C