What is the integral of 12x^2sqrt(1-x^2) ?

The integral of 12x^2sqrt(1-x^2) is 3/2 (arcsin(x)-1/4 sin(4arcsin(x)))+C

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integral of 12x^2sqrt(1-x^2)

\int 12 x^{2} 1 - x^{2} d x

\frac{3}{2} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))) + C

Solution steps

\int 12 x^{2} 1 - x^{2} d x

Take the constant out:

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

= 12 \cdot \int x^{2} 1 - x^{2} d x

Apply Trigonometric Substitution

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

= 12 \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 12 \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

= 12 \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

= 12 \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)

= 12 \cdot \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

u = arcsin (x)

= 12 \cdot \frac{1}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

Simplify

12 \cdot \frac{1}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))) : \frac{3}{2} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

= \frac{3}{2} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

Add a constant to the solution

= \frac{3}{2} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))) + C

\int \frac{x + 5}{9 - ( x - 3 ) ^{2}} d x

\int_{- 1}^{2} \frac{5 y}{y ^{2} - y - 6} d y

f (x) = x^{3} - 6 x^{2} + 9 x + 5

f (x) = x^{2} - 3 x + 6, at x = 2

7 x^{2}, x^{2} + 4

What is the integral of 12x^2sqrt(1-x^2) ?
The integral of 12x^2sqrt(1-x^2) is 3/2 (arcsin(x)-1/4 sin(4arcsin(x)))+C