What is the integral of 1/(sqrt(19-x^2)) ?

The integral of 1/(sqrt(19-x^2)) is arcsin(1/(sqrt(19))x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(19-x^2))

\int \frac{1}{19 - x ^{2}} d x

arcsin (\frac{1}{19} x) + C

Solution steps

\int \frac{1}{19 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = arcsin (\frac{1}{19} x)

= 1 \cdot arcsin (\frac{1}{19} x)

Simplify

= arcsin (\frac{1}{19} x)

Add a constant to the solution

= arcsin (\frac{1}{19} x) + C

3 t^{2} + 13 t

x \to 0 lim (\frac{( x ^{2} + 5 x )}{x})

y = \frac{3}{x ^{2} - 5}

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

x \to 0 - lim (- x + 5)

What is the integral of 1/(sqrt(19-x^2)) ?
The integral of 1/(sqrt(19-x^2)) is arcsin(1/(sqrt(19))x)+C