What is the integral of 2/(sqrt(1-4x^2)) ?

The integral of 2/(sqrt(1-4x^2)) is arcsin(2x)+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(1-4x^2))

\int \frac{2}{1 - 4 x ^{2}} d x

arcsin (2 x) + C

Solution steps

\int \frac{2}{1 - 4 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}{1 - 4 x ^{2}} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

u = arcsin (2 x)

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

Simplify

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

= arcsin (2 x)

Add a constant to the solution

= arcsin (2 x) + C

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 x) d x

\frac{d}{d x} (2 x^{2} - 3 x + 4)

\int \frac{x}{x ^{2} - 4 x + 5} d x

f (x) = \frac{x ^{4}}{2}

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

What is the integral of 2/(sqrt(1-4x^2)) ?
The integral of 2/(sqrt(1-4x^2)) is arcsin(2x)+C