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

The integral of (sqrt(1-x^2))^3 is x(1-x^2)^{3/2}+3/8 (arcsin(x)-1/4 sin(4arcsin(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(1-x^2))^3

\int (1 - x^{2})^{3} d x

x (1 - x^{2})^{\frac{3}{2}} + \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))) + C

Solution steps

\int (1 - x^{2})^{3} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{2} = 2^{- 1}

= \int ((1 - x^{2})^{2^{- 1}})^{3} d x

Simplify

((1 - x^{2})^{2^{- 1}})^{3} : (1 - x^{2})^{\frac{3}{2}}

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

Apply Integration By Parts

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

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

\int - 3 x^{2} 1 - x^{2} d x = - \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

= x (1 - x^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))))

Simplify

= x (1 - x^{2})^{\frac{3}{2}} + \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

Add a constant to the solution

= x (1 - x^{2})^{\frac{3}{2}} + \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))) + C

\int x 4 e^{x^{2}} d x

\frac{1 + x + x ^{2}}{x}

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

\int \frac{( x ^{2} )}{2} - 8 x + 30 d x

\int e^{x} + 12 x^{2} - 4 d x

What is the integral of (sqrt(1-x^2))^3 ?
The integral of (sqrt(1-x^2))^3 is x(1-x^2)^{3/2}+3/8 (arcsin(x)-1/4 sin(4arcsin(x)))+C