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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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