What is the integral of (arcsin(7x))/(sqrt(1-49x^2)) ?

The integral of (arcsin(7x))/(sqrt(1-49x^2)) is 1/14 arcsin^2(7x)+C

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integral of (arcsin(7x))/(sqrt(1-49x^2))

\int \frac{arcsin ( 7 x )}{1 - 49 x ^{2}} d x

\frac{1}{14} arcsin^{2} (7 x) + C

Solution steps

\int \frac{arcsin ( 7 x )}{1 - 49 x ^{2}} d x

Apply u-substitution

: \int \frac{arcsin ( u )}{7 1 - u ^{2}} d u

= \int \frac{arcsin ( u )}{7 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}{7} \cdot \int \frac{arcsin ( u )}{1 - u ^{2}} d u

Apply u-substitution

: \int v d v

= \frac{1}{7} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{7} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{7} \cdot \frac{arcsin ^{2} ( 7 x )}{2}

Simplify

\frac{1}{7} \cdot \frac{arcsin ^{2} ( 7 x )}{2} : \frac{1}{14} arcsin^{2} (7 x)

= \frac{1}{14} arcsin^{2} (7 x)

Add a constant to the solution

= \frac{1}{14} arcsin^{2} (7 x) + C

ln (\frac{5}{x ^{2}})

x^{2} y = 4, x = 2, y = 4

9 cos (- x)

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

\frac{d y}{d x} = \frac{y}{( x + 1 )} + 4 x^{2} + 4 x

What is the integral of (arcsin(7x))/(sqrt(1-49x^2)) ?
The integral of (arcsin(7x))/(sqrt(1-49x^2)) is 1/14 arcsin^2(7x)+C