What is the integral of 1/(sqrt(25-y^2)) ?

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

\int \frac{1}{25 - y ^{2}} d y

arcsin (\frac{1}{5} y) + C

Solution steps

\int \frac{1}{25 - y ^{2}} d y

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}{5} y)

= 1 \cdot arcsin (\frac{1}{5} y)

Simplify

= arcsin (\frac{1}{5} y)

Add a constant to the solution

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

\int - 3 t^{2} d t

\int \frac{y}{5 - y} d y

\int \frac{3 x ^{2} + x - 2}{x ^{4} + 8 x ^{2} + 16} d x

\int x^{3} tan (x^{2}) d x

\int 8 x y^{2} d x

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