What is the integral of sqrt(x^2-x^4) ?

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

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integral of sqrt(x^2-x^4)

\int x^{2} - x^{4} d x

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

Solution steps

\int x^{2} - x^{4} d x

Simplify

x^{2} - x^{4} : x 1 - x^{2}

= \int x 1 - x^{2} d x

Apply u-substitution

: \int - u^{2} d u

= \int - u^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - \frac{u ^{3}}{3}

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

(x, y) \to (0, 0) lim (\frac{x ^{10} - y ^{10}}{x ^{5} + y ^{5}})

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 9 e^{x}

y^{^{'}} + 2 y = x

\int - 5 x^{2} d x

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

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