What is the integral of sqrt(576-16x^2) ?

The integral of sqrt(576-16x^2) is 72(arcsin(x/6)+1/2 sin(2arcsin(x/6)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(576-16x^2)

\int 576 - 16 x^{2} d x

72 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6}))) + C

Solution steps

\int 576 - 16 x^{2} d x

Apply Trigonometric Substitution

: \int 144 cos^{2} (u) d u

= \int 144 cos^{2} (u) d u

Take the constant out:

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

= 144 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 144 \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

= 144 \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

= 144 \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)

= 144 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{4}{24} x)

= 144 \cdot \frac{1}{2} (arcsin (\frac{4}{24} x) + \frac{1}{2} sin (2 arcsin (\frac{4}{24} x)))

Simplify

144 \cdot \frac{1}{2} (arcsin (\frac{4}{24} x) + \frac{1}{2} sin (2 arcsin (\frac{4}{24} x))) : 72 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

= 72 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

Add a constant to the solution

= 72 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6}))) + C

(2 y^{2} + 2 y + 4 x^{2}) d x + (2 x y + x) d y = 0

2 - \frac{3}{x ^{2}}, at x = 1

f (x) = \frac{7}{x}, (1, 7)

x \to 0 + lim (x - ln (x))

\frac{\partial}{\partial x} (\frac{x ^{3}}{1 - x ^{2}})

What is the integral of sqrt(576-16x^2) ?
The integral of sqrt(576-16x^2) is 72(arcsin(x/6)+1/2 sin(2arcsin(x/6)))+C