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

The integral of+x^2sqrt(a^2-x^2) is 1/8 a^4(arcsin(x/a)-1/4 sin(4arcsin(x/a)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of+x^2sqrt(a^2-x^2)

\int + x^{2} a^{2} - x^{2} d x

\frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a}))) + C

Solution steps

\int x^{2} a^{2} - x^{2} d x

Apply Trigonometric Substitution

: \int a^{4} sin^{2} (u) cos^{2} (u) d u

= \int a^{4} sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= a^{4} \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= a^{4} \cdot \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= a^{4} \frac{1}{8} \cdot \int 1 - cos (4 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

= a^{4} \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= a^{4} \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

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

= a^{4} \frac{1}{8} (arcsin (\frac{1}{a} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{a} x)))

Simplify

a^{4} \frac{1}{8} (arcsin (\frac{1}{a} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{a} x))) : \frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a})))

= \frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a})))

Add a constant to the solution

= \frac{1}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a}))) + C

\int \frac{e ^{9 t}}{t} d t

arcsin (x) + 1 + x^{2}

\int_{0}^{5} (25 - x^{2}) d x

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

\int_{0}^{π} 9 x^{2} sin (\frac{1}{2} x) d x

What is the integral of+x^2sqrt(a^2-x^2) ?
The integral of+x^2sqrt(a^2-x^2) is 1/8 a^4(arcsin(x/a)-1/4 sin(4arcsin(x/a)))+C