What is the integral of x^3(1-4x^2)^{5/2} ?

The integral of x^3(1-4x^2)^{5/2} is -1/504 (14x^2+1)(-4x^2+1)^{7/2}+C

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integral of x^3(1-4x^2)^{5/2}

\int x^{3} (1 - 4 x^{2})^{\frac{5}{2}} d x

- \frac{1}{504} (14 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{7}{2}} + C

Solution steps

\int x^{3} (1 - 4 x^{2})^{\frac{5}{2}} d x

Apply u-substitution

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

Expand

u^{\frac{5}{2}} (u - 1) : u^{\frac{7}{2}} - u^{\frac{5}{2}}

= \frac{1}{32} \cdot \int u^{\frac{7}{2}} - u^{\frac{5}{2}} 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}{32} (\int u^{\frac{7}{2}} d u - \int u^{\frac{5}{2}} d u)

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

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

= \frac{1}{32} (\frac{2}{9} u^{\frac{9}{2}} - \frac{2}{7} u^{\frac{7}{2}})

Substitute back

u = 1 - 4 x^{2}

= \frac{1}{32} (\frac{2}{9} (1 - 4 x^{2})^{\frac{9}{2}} - \frac{2}{7} (1 - 4 x^{2})^{\frac{7}{2}})

Simplify

\frac{1}{32} (\frac{2}{9} (1 - 4 x^{2})^{\frac{9}{2}} - \frac{2}{7} (1 - 4 x^{2})^{\frac{7}{2}}) : - \frac{1}{504} (14 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{7}{2}}

= - \frac{1}{504} (14 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{7}{2}}

Add a constant to the solution

= - \frac{1}{504} (14 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{7}{2}} + C

What is the integral of x^3(1-4x^2)^{5/2} ?
The integral of x^3(1-4x^2)^{5/2} is -1/504 (14x^2+1)(-4x^2+1)^{7/2}+C