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

The integral of (x^3)/(3sqrt(1-x^2)) is 1/3 (1/3 (1-x^2)^{3/2}-sqrt(1-x^2))+C

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

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

\frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2}) + C

Solution steps

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

Apply u-substitution

: \int u^{2} - 1 d u

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

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

\int 1 d u = u

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x + x, at x = 36

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