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

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

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

integral

y (1 - y^{2})^{\frac{1}{2}}

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

2 e^{- x}

\frac{1}{( 1 + x ^{2} ) ^{0.5}}

\frac{- 1}{y + 1}

x^{2} e^{2} x

\frac{3}{x ^{4}}

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