What is the integral of ((3x^2)/(sqrt(64-x^2))) ?

The integral of ((3x^2)/(sqrt(64-x^2))) is 96(arcsin(1/8 x)-1/2 sin(2arcsin(1/8 x)))+C

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integral of ((3x^2)/(sqrt(64-x^2)))

\int (\frac{3 x ^{2}}{64 - x ^{2}}) d x

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

Solution steps

\int \frac{3 x ^{2}}{64 - x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{2}}{64 - x ^{2}} d x

Apply Trigonometric Substitution

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

= 3 \cdot \int 64 sin^{2} (u) d u

Take the constant out:

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

= 3 \cdot 64 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 3 \cdot 64 \cdot \int \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= 3 \cdot 64 \cdot \frac{1}{2} \cdot \int 1 - cos (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

= 3 \cdot 64 \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 3 \cdot 64 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{8} x)

= 3 \cdot 64 \cdot \frac{1}{2} (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x)))

Simplify

3 \cdot 64 \cdot \frac{1}{2} (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x))) : 96 (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x)))

= 96 (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x)))

Add a constant to the solution

= 96 (arcsin (\frac{1}{8} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{8} x))) + C

x \to 0 + lim (\frac{x ^{- 2}}{x})

\int x 36 - x^{2} d x

\int \frac{e ^{t}}{( 1 + e ^{(2 t)} ) ^{(\frac{3}{2})}} d t

\int 3 x - 2 d x

\frac{d y}{d x} = 3 y \cdot e^{x + 3}

What is the integral of ((3x^2)/(sqrt(64-x^2))) ?
The integral of ((3x^2)/(sqrt(64-x^2))) is 96(arcsin(1/8 x)-1/2 sin(2arcsin(1/8 x)))+C