What is the integral of sqrt(900-25x^2) ?

The integral of sqrt(900-25x^2) is 90(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(900-25x^2)

\int 900 - 25 x^{2} d x

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

Solution steps

\int 900 - 25 x^{2} d x

Apply Trigonometric Substitution

: \int 180 cos^{2} (u) d u

= \int 180 cos^{2} (u) d u

Take the constant out:

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

= 180 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 180 \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

= 180 \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

= 180 \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)

= 180 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{5}{30} x)

= 180 \cdot \frac{1}{2} (arcsin (\frac{5}{30} x) + \frac{1}{2} sin (2 arcsin (\frac{5}{30} x)))

Simplify

180 \cdot \frac{1}{2} (arcsin (\frac{5}{30} x) + \frac{1}{2} sin (2 arcsin (\frac{5}{30} x))) : 90 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

= 90 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

Add a constant to the solution

= 90 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6}))) + C

\int 4 x e^{\frac{4 x}{3}} d x

\int x (2 x - 1) d x

n = 2 \sum \infty \frac{n + 2}{2 n + 3}

x \to 1 lim (x (2))

\frac{d}{d x} (a x^{2} + b y (x)^{2} + λ (x^{2} + y^{2} - c^{2}))

What is the integral of sqrt(900-25x^2) ?
The integral of sqrt(900-25x^2) is 90(arcsin(x/6)+1/2 sin(2arcsin(x/6)))+C