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

The integral of x(1-x^2)^{1/2} is 1/2 x^2(1-x^2)^{1/2}+1/6 ((1-x^2)^{3/2}-3sqrt(1-x^2))+C

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

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

\frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} + \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2}) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{x ^{3}}{2 ( 1 - x ^{2} ) ^{\frac{1}{2}}} d x = - \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2})

= \frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} - (- \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2}))

Simplify

= \frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} + \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2})

Add a constant to the solution

= \frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} + \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2}) + C

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What is the integral of x(1-x^2)^{1/2} ?
The integral of x(1-x^2)^{1/2} is 1/2 x^2(1-x^2)^{1/2}+1/6 ((1-x^2)^{3/2}-3sqrt(1-x^2))+C