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

The integral of sqrt(t^2-2t^4) is -1/6 (1-2t^2)^{3/2}+C

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

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

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

Solution steps

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

Simplify

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

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

t^{2} = (t),

assuming

t \geq 0

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

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 - 2 t^{2}

= - \frac{1}{4} \cdot \frac{2}{3} (1 - 2 t^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (1 - 2 t^{2})^{\frac{3}{2}} : - \frac{1}{6} (1 - 2 t^{2})^{\frac{3}{2}}

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

Add a constant to the solution

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

f (x) = \frac{4 x + 3}{4 x - 3}

\int_{- 3}^{3} \frac{16}{x ^{2} - 6 x - 55} d x

\frac{d}{d x} ((1 - e^{x}) (x + e^{x}))

\frac{\partial}{\partial z} (x^{2} y z^{2} + sin (yz))

\int 1 + \frac{x ^{2}}{2 - x ^{2}} d x

What is the integral of sqrt(t^2-2t^4) ?
The integral of sqrt(t^2-2t^4) is -1/6 (1-2t^2)^{3/2}+C