What is the integral of 6xarccos(6x) ?

The integral of 6xarccos(6x) is 1/6 (18x^2arccos(6x)+1/4 (arcsin(6x)-1/2 sin(2arcsin(6x))))+C

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integral of 6xarccos(6x)

\int 6 x arccos (6 x) d x

\frac{1}{6} (18 x^{2} arccos (6 x) + \frac{1}{4} (arcsin (6 x) - \frac{1}{2} sin (2 arcsin (6 x)))) + C

Solution steps

\int 6 x arccos (6 x) d x

Take the constant out:

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

= 6 \cdot \int x arccos (6 x) d x

Apply u-substitution

: \int \frac{u arccos ( u )}{36} d u

= 6 \cdot \int \frac{u arccos ( u )}{36} d u

Take the constant out:

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

= 6 \cdot \frac{1}{36} \cdot \int u arccos (u) d u

Apply Integration By Parts

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

= 6 \cdot \frac{1}{36} (\frac{1}{2} u^{2} arccos (u) - \int - \frac{u ^{2}}{2 1 - u ^{2}} d u)

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

= 6 \cdot \frac{1}{36} (\frac{1}{2} u^{2} arccos (u) - (- \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))))

Substitute back

u = 6 x

= 6 \cdot \frac{1}{36} (\frac{1}{2} (6 x)^{2} arccos (6 x) - (- \frac{1}{4} (arcsin (6 x) - \frac{1}{2} sin (2 arcsin (6 x)))))

Simplify

6 \cdot \frac{1}{36} (\frac{1}{2} (6 x)^{2} arccos (6 x) - (- \frac{1}{4} (arcsin (6 x) - \frac{1}{2} sin (2 arcsin (6 x))))) : \frac{1}{6} (18 x^{2} arccos (6 x) + \frac{1}{4} (arcsin (6 x) - \frac{1}{2} sin (2 arcsin (6 x))))

= \frac{1}{6} (18 x^{2} arccos (6 x) + \frac{1}{4} (arcsin (6 x) - \frac{1}{2} sin (2 arcsin (6 x))))

Add a constant to the solution

= \frac{1}{6} (18 x^{2} arccos (6 x) + \frac{1}{4} (arcsin (6 x) - \frac{1}{2} sin (2 arcsin (6 x)))) + C

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What is the integral of 6xarccos(6x) ?
The integral of 6xarccos(6x) is 1/6 (18x^2arccos(6x)+1/4 (arcsin(6x)-1/2 sin(2arcsin(6x))))+C