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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{4} arcsin^{2} (2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{arcsin ( u )}{2 1 - u ^{2}} d u

= \int \frac{arcsin ( u )}{2 1 - u ^{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 \frac{arcsin ( u )}{1 - u ^{2}} d u

Apply u-substitution

: \int v d v

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

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{arcsin ^{2} ( 2 x )}{2} : \frac{1}{4} arcsin^{2} (2 x)

= \frac{1}{4} arcsin^{2} (2 x)

Add a constant to the solution

= \frac{1}{4} arcsin^{2} (2 x) + C

f (x) = x + 5, at x = - 4

\int tan^{2} (5 x) sec^{4} (5 x) d x

\frac{d}{d t} (\frac{( t + 1 ) ^{2}}{( t - 1 ) ^{2}})

\int sin (wt - b x) d t

\frac{\partial}{\partial y} (2 x^{2} + 6 x y + y^{2})

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