What is the integral of 1/(sqrt(400-x^2)) ?

The integral of 1/(sqrt(400-x^2)) is arcsin(1/20 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(400-x^2))

\int \frac{1}{400 - x ^{2}} d x

arcsin (\frac{1}{20} x) + C

Solution steps

\int \frac{1}{400 - 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}{20} x)

= 1 \cdot arcsin (\frac{1}{20} x)

Simplify

= arcsin (\frac{1}{20} x)

Add a constant to the solution

= arcsin (\frac{1}{20} x) + C

t (x) v^{''} + v^{'} = 0

\int \frac{3}{x ln ( 5 x )} d x

y^{10} \frac{d y}{d x} = - 9 cos (π x)

\int sin^{2} (θ) cos^{3} (θ) d θ

y = (10 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

What is the integral of 1/(sqrt(400-x^2)) ?
The integral of 1/(sqrt(400-x^2)) is arcsin(1/20 x)+C