What is the integral of 1/(sqrt(20-x^2)) ?

The integral of 1/(sqrt(20-x^2)) is arcsin(1/(2sqrt(5))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(20-x^2))

\int \frac{1}{20 - x ^{2}} d x

arcsin (\frac{1}{2 5} x) + C

Solution steps

\int \frac{1}{20 - 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}{2 5} x)

= 1 \cdot arcsin (\frac{1}{2 5} x)

Simplify

= arcsin (\frac{1}{2 5} x)

Add a constant to the solution

= arcsin (\frac{1}{2 5} x) + C

\int \frac{2 ln ^{2} ( x ) + 16}{x} d x

\frac{d}{d y} (2)

\frac{\partial}{\partial x} (n y ln (n x))

\int 4 t + 2 d t

\frac{\partial}{\partial x} (ln (3 x + x^{2} y^{2}))

What is the integral of 1/(sqrt(20-x^2)) ?
The integral of 1/(sqrt(20-x^2)) is arcsin(1/(2sqrt(5))x)+C