What is the integral of 1/(sqrt(100-x^2)) ?

The integral of 1/(sqrt(100-x^2)) is arcsin(1/10 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(100-x^2))

\int \frac{1}{100 - x ^{2}} d x

arcsin (\frac{1}{10} x) + C

Solution steps

\int \frac{1}{100 - 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}{10} x)

= 1 \cdot arcsin (\frac{1}{10} x)

Simplify

= arcsin (\frac{1}{10} x)

Add a constant to the solution

= arcsin (\frac{1}{10} x) + C

\frac{\partial}{\partial x} (\frac{- y ^{2}}{( x - y ) ^{2}})

\int e^{i 7 x} cos (5 x) d x

\frac{\partial}{\partial x} (5 e^{x - y})

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

\int \frac{2 u}{1 + u ^{2}} d u

What is the integral of 1/(sqrt(100-x^2)) ?
The integral of 1/(sqrt(100-x^2)) is arcsin(1/10 x)+C