What is the integral of sqrt(225-x^2) ?

The integral of sqrt(225-x^2) is 225/2 (arcsin(1/15 x)+1/2 sin(2arcsin(1/15 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(225-x^2)

\int 225 - x^{2} d x

\frac{225}{2} (arcsin (\frac{1}{15} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{15} x))) + C

Solution steps

\int 225 - x^{2} d x

Apply Trigonometric Substitution

: \int 225 cos^{2} (u) d u

= \int 225 cos^{2} (u) d u

Take the constant out:

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

= 225 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 225 \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

= 225 \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

= 225 \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)

= 225 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{15} x)

= 225 \cdot \frac{1}{2} (arcsin (\frac{1}{15} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{15} x)))

Simplify

225 \cdot \frac{1}{2} (arcsin (\frac{1}{15} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{15} x))) : \frac{225}{2} (arcsin (\frac{1}{15} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{15} x)))

= \frac{225}{2} (arcsin (\frac{1}{15} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{15} x)))

Add a constant to the solution

= \frac{225}{2} (arcsin (\frac{1}{15} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{15} x))) + C

x \to 4 lim (5 x - 4)

\int (x + 1) e^{(x + 1)^{2}} d x

y = (3 x + 2) x^{2}

\frac{d}{d x} (2 x^{2} + y^{2} - ln (x y^{2}))

y = 2 + \frac{1}{x}, (1, 3)

What is the integral of sqrt(225-x^2) ?
The integral of sqrt(225-x^2) is 225/2 (arcsin(1/15 x)+1/2 sin(2arcsin(1/15 x)))+C