What is the integral of y^7sqrt(1-y^4) ?

The integral of y^7sqrt(1-y^4) is -1/30 (3y^4+2)(-y^4+1)^{3/2}+C

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integral of y^7sqrt(1-y^4)

\int y^{7} 1 - y^{4} d y

- \frac{1}{30} (3 y^{4} + 2) (- y^{4} + 1)^{\frac{3}{2}} + C

Solution steps

\int y^{7} 1 - y^{4} d y

Apply u-substitution

: \int - \frac{( - u + 1 ) u}{4} d u

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

Expand

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

= - \frac{1}{4} \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}{4} (- \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}{4} (- \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - y^{4}

= - \frac{1}{4} (- \frac{2}{5} (1 - y^{4})^{\frac{5}{2}} + \frac{2}{3} (1 - y^{4})^{\frac{3}{2}})

Simplify

- \frac{1}{4} (- \frac{2}{5} (1 - y^{4})^{\frac{5}{2}} + \frac{2}{3} (1 - y^{4})^{\frac{3}{2}}) : - \frac{1}{30} (3 y^{4} + 2) (- y^{4} + 1)^{\frac{3}{2}}

= - \frac{1}{30} (3 y^{4} + 2) (- y^{4} + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{30} (3 y^{4} + 2) (- y^{4} + 1)^{\frac{3}{2}} + C

\frac{d}{d x} (- x e^{2 x})

x \cdot y^{2} \cdot \frac{d y}{d x} = y^{3} - 4 \cdot x^{3}, y (1) = 2

x^{2} - 8, 7 x

x \to 0 lim (\frac{sin ( 5 x )}{5})

y^{^{''}} - 2 y^{^{'}} - 8 y = x e^{x} + 4

What is the integral of y^7sqrt(1-y^4) ?
The integral of y^7sqrt(1-y^4) is -1/30 (3y^4+2)(-y^4+1)^{3/2}+C