What is the integral of sqrt(400-x^2) ?

The integral of sqrt(400-x^2) is 200(arcsin(1/20 x)+1/2 sin(2arcsin(1/20 x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(400-x^2)

\int 400 - x^{2} d x

200 (arcsin (\frac{1}{20} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{20} x))) + C

Solution steps

\int 400 - x^{2} d x

Apply Trigonometric Substitution

: \int 400 cos^{2} (u) d u

= \int 400 cos^{2} (u) d u

Take the constant out:

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

= 400 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 400 \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

= 400 \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

= 400 \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)

= 400 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{20} x)

= 400 \cdot \frac{1}{2} (arcsin (\frac{1}{20} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{20} x)))

Simplify

400 \cdot \frac{1}{2} (arcsin (\frac{1}{20} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{20} x))) : 200 (arcsin (\frac{1}{20} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{20} x)))

= 200 (arcsin (\frac{1}{20} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{20} x)))

Add a constant to the solution

= 200 (arcsin (\frac{1}{20} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{20} x))) + C

5 cos (3 x), 5 sin (6 x), [0, \frac{π}{6}]

cos (t + π)

2 x^{3} + 3 x^{2} - 12 x + 9

x \to 0 + lim ((1 - x)^{\frac{1}{x}})

\frac{d y}{d x} = 3 x^{2} + 1

What is the integral of sqrt(400-x^2) ?
The integral of sqrt(400-x^2) is 200(arcsin(1/20 x)+1/2 sin(2arcsin(1/20 x)))+C