What is the integral of sqrt(t^2-t^4) 1 ?

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

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integral of sqrt(t^2-t^4) 1

\int t^{2} - t^{4} \frac{d}{1} d t

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

Solution steps

\int t^{2} - t^{4} \frac{d}{1} d t

Take the constant out:

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

= d \cdot \int t^{2} - t^{4} d t

Simplify

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

= d \cdot \int t 1 - t^{2} d t

Apply u-substitution

: \int - u^{2} d u

= d \cdot \int - u^{2} d u

Take the constant out:

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

= d (- \int u^{2} d u)

Apply the Power Rule

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

= d (- \frac{u ^{3}}{3})

Substitute back

u = 1 - t^{2}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'''}} + 2 y^{^{''}} + 2 y^{^{'}} = 0

arctan (\frac{y}{x})

(sin (t))^{^{'}}

f (x) = 4 x^{2} - 15

e^{2 x} y^{^{'}} - 4 y = 0

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