What is the integral of arccos(sqrt(x)) ?

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

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integral of arccos(sqrt(x))

\int arccos (x) d x

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

Solution steps

\int arccos (x) d x

Apply u-substitution

= \int arccos (u) \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int arccos (u) u d u

Apply Integration By Parts

= 2 (\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)))

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

Substitute back

u = x

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

Simplify

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

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

Add a constant to the solution

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

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