What is the integral of (arcsin(4x))/(sqrt(1-16x^2)) ?

The integral of (arcsin(4x))/(sqrt(1-16x^2)) is 1/8 arcsin^2(4x)+C

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integral of (arcsin(4x))/(sqrt(1-16x^2))

\int \frac{arcsin ( 4 x )}{1 - 16 x ^{2}} d x

\frac{1}{8} arcsin^{2} (4 x) + C

Solution steps

\int \frac{arcsin ( 4 x )}{1 - 16 x ^{2}} d x

Apply u-substitution

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

= \int \frac{arcsin ( u )}{4 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}{4} \cdot \int \frac{arcsin ( u )}{1 - u ^{2}} d u

Apply u-substitution

: \int v d v

= \frac{1}{4} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

\frac{1}{4} \cdot \frac{arcsin ^{2} ( 4 x )}{2} : \frac{1}{8} arcsin^{2} (4 x)

= \frac{1}{8} arcsin^{2} (4 x)

Add a constant to the solution

= \frac{1}{8} arcsin^{2} (4 x) + C

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What is the integral of (arcsin(4x))/(sqrt(1-16x^2)) ?
The integral of (arcsin(4x))/(sqrt(1-16x^2)) is 1/8 arcsin^2(4x)+C