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

The integral of x^2(4-x^2)^{1/2} is 1/4 x^3(-x^2+4)^{1/2}+2arcsin(1/2 x)-sin(2arcsin(1/2 x))+C

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

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

\frac{1}{4} x^{3} (- x^{2} + 4)^{\frac{1}{2}} + 2 arcsin (\frac{1}{2} x) - sin (2 arcsin (\frac{1}{2} x)) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{x ^{4}}{3 ( 4 - x ^{2} ) ^{\frac{1}{2}}} d x = \frac{1}{12} x^{3} (- x^{2} + 4)^{\frac{1}{2}} - 2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x))

= \frac{1}{3} x^{3} (4 - x^{2})^{\frac{1}{2}} - (\frac{1}{12} x^{3} (- x^{2} + 4)^{\frac{1}{2}} - 2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x)))

Simplify

\frac{1}{3} x^{3} (4 - x^{2})^{\frac{1}{2}} - (\frac{1}{12} x^{3} (- x^{2} + 4)^{\frac{1}{2}} - 2 arcsin (\frac{1}{2} x) + sin (2 arcsin (\frac{1}{2} x))) : \frac{1}{4} x^{3} (- x^{2} + 4)^{\frac{1}{2}} + 2 arcsin (\frac{1}{2} x) - sin (2 arcsin (\frac{1}{2} x))

= \frac{1}{4} x^{3} (- x^{2} + 4)^{\frac{1}{2}} + 2 arcsin (\frac{1}{2} x) - sin (2 arcsin (\frac{1}{2} x))

Add a constant to the solution

= \frac{1}{4} x^{3} (- x^{2} + 4)^{\frac{1}{2}} + 2 arcsin (\frac{1}{2} x) - sin (2 arcsin (\frac{1}{2} x)) + C

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What is the integral of x^2(4-x^2)^{1/2} ?
The integral of x^2(4-x^2)^{1/2} is 1/4 x^3(-x^2+4)^{1/2}+2arcsin(1/2 x)-sin(2arcsin(1/2 x))+C