What is the integral of x/(sqrt(a^4-x^4)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(sqrt(a^4-x^4))

\int \frac{x}{a ^{4} - x ^{4}} d x

\frac{1}{2} arcsin (\frac{x ^{2}}{a ^{2}}) + C

Solution steps

\int \frac{x}{a ^{4} - x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 a ^{4} - u ^{2}} d u

= \int \frac{1}{2 a ^{4} - u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{a ^{4} - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{a ^{2}} x^{2})

Simplify

\frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{a ^{2}} x^{2}) : \frac{1}{2} arcsin (\frac{x ^{2}}{a ^{2}})

= \frac{1}{2} arcsin (\frac{x ^{2}}{a ^{2}})

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{x ^{2}}{a ^{2}}) + C

y = 5 x - x^{2}

x \to \infty lim (2 x + \frac{1}{x})

\int_{\frac{3}{2}}^{3} 9 - x^{2} d x

\frac{d}{d x} (cos (x^{2}))

\int_{0}^{3} \frac{1}{x ^{2} - x - 2} d x

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