What is the integral of 1/(sqrt(8-2x^2)) ?

The integral of 1/(sqrt(8-2x^2)) is 1/(sqrt(2))arcsin(x/2)+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(8-2x^2))

\int \frac{1}{8 - 2 x ^{2}} d x

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

Solution steps

\int \frac{1}{8 - 2 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{2}{2 2} x)

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

Simplify

\frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{2 2} x) : \frac{1}{2} arcsin (\frac{x}{2})

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

Add a constant to the solution

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

\int \frac{2 x}{x ^{2} + 6 x + 5} d x

3 t cos (3 t)

x \to 1 lim (e^{x - 2})

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

y^{^{''}} + 16 y = 4 sec (4 t)

What is the integral of 1/(sqrt(8-2x^2)) ?
The integral of 1/(sqrt(8-2x^2)) is 1/(sqrt(2))arcsin(x/2)+C