What is the integral of sqrt(25-y^2) ?

The integral of sqrt(25-y^2) is 25/2 (arcsin(1/5 y)+1/2 sin(2arcsin(1/5 y)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(25-y^2)

\int 25 - y^{2} d y

\frac{25}{2} (arcsin (\frac{1}{5} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} y))) + C

Solution steps

\int 25 - y^{2} d y

Apply Trigonometric Substitution

: \int 25 cos^{2} (u) d u

= \int 25 cos^{2} (u) d u

Take the constant out:

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

= 25 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 25 \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

= 25 \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

= 25 \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)

= 25 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{5} y)

= 25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} y)))

Simplify

25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} y))) : \frac{25}{2} (arcsin (\frac{1}{5} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} y)))

= \frac{25}{2} (arcsin (\frac{1}{5} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} y)))

Add a constant to the solution

= \frac{25}{2} (arcsin (\frac{1}{5} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} y))) + C

3 x - 6, - 2 x + 10, 2, 5

\frac{d y}{d x} + x y = 0

\int \frac{x}{x ^{2} - 5 x + 6} d x

\frac{d}{d x} (1 + x ln (2))

\int_{1}^{4} (x - 1) d x

What is the integral of sqrt(25-y^2) ?
The integral of sqrt(25-y^2) is 25/2 (arcsin(1/5 y)+1/2 sin(2arcsin(1/5 y)))+C