What is the integral of arcsin(bx) ?

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

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integral of arcsin(bx)

\int arcsin (b x) d x

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

Solution steps

\int arcsin (b x) d x

Apply u-substitution

: \int \frac{arcsin ( u )}{b} d u

= \int \frac{arcsin ( u )}{b} d u

Take the constant out:

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

= \frac{1}{b} \cdot \int arcsin (u) d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = b x

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

Simplify

\frac{1}{b} (b x arcsin (b x) - (- 1 - (b x)^{2})) : x arcsin (b x) + \frac{1 - b ^{2} x ^{2}}{b}

= x arcsin (b x) + \frac{1 - b ^{2} x ^{2}}{b}

Add a constant to the solution

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

y \frac{d y}{d x} = sin (x)

\frac{d y}{d x} = x 1 - y^{2}

\frac{d}{d y} (y^{2} + y)

\int \frac{3 x}{1 - 9 x ^{4}} d x

\int \frac{\frac{1}{120}}{- P + 120} d P

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