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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 16 sin^{2} (u) cos^{2} (u) d u

= \int 16 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

= 16 \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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)

= 16 \cdot \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

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

= 16 \cdot \frac{1}{8} (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

Simplify

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

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

Add a constant to the solution

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

\int (z + 1) e^{3 z} d z

y^{3} + 3 y^{2} = x^{4} - 3 x^{2}, (- 2, 1)

\int 3 x (x - 2) d x

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

\int \frac{cos ( x )}{1 - cos ( x )} d x

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