What is the integral of 2/(sqrt(25-x^2)) ?

The integral of 2/(sqrt(25-x^2)) is 2arcsin(1/5 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/(sqrt(25-x^2))

\int \frac{2}{25 - x ^{2}} d x

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

Solution steps

\int \frac{2}{25 - x ^{2}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 2 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

\int cos (12 x) cos (3 x) d x

\int - x^{3} sin (x) d x

\int \frac{x ^{3}}{x ^{2} - 64} d x

\int (csc^{2} (x))^{2} csc (x) d x

\int csch (2 z) d z

What is the integral of 2/(sqrt(25-x^2)) ?
The integral of 2/(sqrt(25-x^2)) is 2arcsin(1/5 x)+C