What is the integral of (arcsin(6x))/(sqrt(1-36x^2)) ?

The integral of (arcsin(6x))/(sqrt(1-36x^2)) is 1/12 arcsin^2(6x)+C

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integral of (arcsin(6x))/(sqrt(1-36x^2))

\int \frac{arcsin ( 6 x )}{1 - 36 x ^{2}} d x

\frac{1}{12} arcsin^{2} (6 x) + C

Solution steps

\int \frac{arcsin ( 6 x )}{1 - 36 x ^{2}} d x

Apply u-substitution

: \int \frac{arcsin ( u )}{6 1 - u ^{2}} d u

= \int \frac{arcsin ( u )}{6 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}{6} \cdot \int \frac{arcsin ( u )}{1 - u ^{2}} d u

Apply u-substitution

: \int v d v

= \frac{1}{6} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{6} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{6} \cdot \frac{arcsin ^{2} ( 6 x )}{2}

Simplify

\frac{1}{6} \cdot \frac{arcsin ^{2} ( 6 x )}{2} : \frac{1}{12} arcsin^{2} (6 x)

= \frac{1}{12} arcsin^{2} (6 x)

Add a constant to the solution

= \frac{1}{12} arcsin^{2} (6 x) + C

y = x^{\frac{1}{x}}

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\int_{0}^{4} x e^{- 0.1 x} d x

\frac{d}{d x} (ln^{2} (x - 1))

What is the integral of (arcsin(6x))/(sqrt(1-36x^2)) ?
The integral of (arcsin(6x))/(sqrt(1-36x^2)) is 1/12 arcsin^2(6x)+C