What is the integral of arccos(ax) ?

The integral of arccos(ax) is xarccos(ax)-(sqrt(1-a^2x^2))/a+C

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integral of arccos(ax)

\int arccos (a x) d x

x arccos (a x) - \frac{1 - a ^{2} x ^{2}}{a} + C

Solution steps

\int arccos (a x) d x

Apply u-substitution

= \int \frac{arccos ( u )}{a} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int arccos (u) d u

Apply Integration By Parts

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

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

= \frac{1}{a} (u arccos (u) - 1 - u^{2})

Substitute back

u = a x

= \frac{1}{a} (a x arccos (a x) - 1 - (a x)^{2})

Simplify

\frac{1}{a} (a x arccos (a x) - 1 - (a x)^{2}) : x arccos (a x) - \frac{1 - a ^{2} x ^{2}}{a}

= x arccos (a x) - \frac{1 - a ^{2} x ^{2}}{a}

Add a constant to the solution

= x arccos (a x) - \frac{1 - a ^{2} x ^{2}}{a} + C

What is the integral of arccos(ax) ?
The integral of arccos(ax) is xarccos(ax)-(sqrt(1-a^2x^2))/a+C