What is the integral of sqrt(x^6-x^8) ?

The integral of sqrt(x^6-x^8) is -1/15 (3x^2+2)(-x^2+1)^{3/2}+C

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integral of sqrt(x^6-x^8)

\int x^{6} - x^{8} d x

- \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{6} - x^{8} d x

Simplify

x^{6} - x^{8} : x^{3} 1 - x^{2}

= \int x^{3} 1 - x^{2} d x

Apply u-substitution

: \int - \frac{( - u + 1 ) u}{2} d u

= \int - \frac{( - u + 1 ) u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int (- u + 1) u d u

Expand

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

= - \frac{1}{2} \cdot \int - u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{2} (- \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - x^{2}

= - \frac{1}{2} (- \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}})

Simplify

- \frac{1}{2} (- \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}}) : - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}}

= - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}} + C

f (x) = sin (sin (x)), (2 π, 0)

\int \frac{I n x}{x} d x

\int \frac{12}{sec ( 2 x ) - 1} d x

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

y^{^{'}} - y = e^{- t} sin (t)

What is the integral of sqrt(x^6-x^8) ?
The integral of sqrt(x^6-x^8) is -1/15 (3x^2+2)(-x^2+1)^{3/2}+C