What is the integral of (arccos(x/2))/(sqrt(4-x^2)) ?

The integral of (arccos(x/2))/(sqrt(4-x^2)) is -(arccos^2(x/2))/2+C

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integral of (arccos(x/2))/(sqrt(4-x^2))

\int \frac{arccos ( \frac{x}{2} )}{4 - x ^{2}} d x

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

Solution steps

\int \frac{arccos ( \frac{x}{2} )}{4 - x ^{2}} d x

Apply u-substitution

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

Apply u-substitution

= \int - v d v

Take the constant out:

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

= - \int v d v

Apply the Power Rule

= - \frac{v ^{2}}{2}

Substitute back

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

Add a constant to the solution

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

What is the integral of (arccos(x/2))/(sqrt(4-x^2)) ?
The integral of (arccos(x/2))/(sqrt(4-x^2)) is -(arccos^2(x/2))/2+C