What is the integral of x^2sqrt(8-x^2) ?

The integral of x^2sqrt(8-x^2) is 8(arcsin(1/(2sqrt(2))x)-1/4 sin(4arcsin(1/(2sqrt(2))x)))+C

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integral of x^2sqrt(8-x^2)

\int x^{2} 8 - x^{2} d x

8 (arcsin (\frac{1}{2 2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2 2} x))) + C

Solution steps

\int x^{2} 8 - x^{2} d x

Apply Trigonometric Substitution

: \int 64 sin^{2} (u) cos^{2} (u) d u

= \int 64 sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= 64 \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 64 \cdot \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= 64 \cdot \frac{1}{8} \cdot \int 1 - cos (4 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

= 64 \cdot \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= 64 \cdot \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

u = arcsin (\frac{1}{2 2} x)

= 64 \cdot \frac{1}{8} (arcsin (\frac{1}{2 2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2 2} x)))

Simplify

64 \cdot \frac{1}{8} (arcsin (\frac{1}{2 2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2 2} x))) : 8 (arcsin (\frac{1}{2 2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2 2} x)))

= 8 (arcsin (\frac{1}{2 2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2 2} x)))

Add a constant to the solution

= 8 (arcsin (\frac{1}{2 2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2 2} x))) + C

\int cot^{2} (x) (1 + tan^{2} (x)) d x

6 t^{4}

f (x) = 4 - x^{2}, at x = - 1

\int \frac{1}{10 x ^{7}} d x

y^{'} = y (3 - y)

What is the integral of x^2sqrt(8-x^2) ?
The integral of x^2sqrt(8-x^2) is 8(arcsin(1/(2sqrt(2))x)-1/4 sin(4arcsin(1/(2sqrt(2))x)))+C