What is the integral of (x^2)/(sqrt(1-3x^2)) ?

The integral of (x^2)/(sqrt(1-3x^2)) is 1/(6sqrt(3))(arcsin(sqrt(3)x)-1/2 sin(2arcsin(sqrt(3)x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(sqrt(1-3x^2))

\int \frac{x ^{2}}{1 - 3 x ^{2}} d x

\frac{1}{6 3} (arcsin (3 x) - \frac{1}{2} sin (2 arcsin (3 x))) + C

Solution steps

\int \frac{x ^{2}}{1 - 3 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sin ^{2} ( u )}{3 3} d u

= \int \frac{sin ^{2} ( u )}{3 3} d u

Take the constant out:

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

= \frac{1}{3 3} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= \frac{1}{3 3} \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

= \frac{1}{3 3} \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

= \frac{1}{3 3} \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)

= \frac{1}{3 3} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (3 x)

= \frac{1}{3 3} \cdot \frac{1}{2} (arcsin (3 x) - \frac{1}{2} sin (2 arcsin (3 x)))

Simplify

\frac{1}{3 3} \cdot \frac{1}{2} (arcsin (3 x) - \frac{1}{2} sin (2 arcsin (3 x))) : \frac{1}{6 3} (arcsin (3 x) - \frac{1}{2} sin (2 arcsin (3 x)))

= \frac{1}{6 3} (arcsin (3 x) - \frac{1}{2} sin (2 arcsin (3 x)))

Add a constant to the solution

= \frac{1}{6 3} (arcsin (3 x) - \frac{1}{2} sin (2 arcsin (3 x))) + C

\int x + \frac{1}{2} d x

\int x (1) d x

\int \frac{sin ( x )}{3 - cos ( x )} d x

\int \frac{1}{4 + 7 x ^{2}} d x

\int \frac{x ^{3} + 4 x ^{2} + 2 x}{x ^{2} + 2} d x

What is the integral of (x^2)/(sqrt(1-3x^2)) ?
The integral of (x^2)/(sqrt(1-3x^2)) is 1/(6sqrt(3))(arcsin(sqrt(3)x)-1/2 sin(2arcsin(sqrt(3)x)))+C