What is the integral of (arccos(x)-x)/(sqrt(1-x^2)) ?

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

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integral of (arccos(x)-x)/(sqrt(1-x^2))

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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\frac{d}{d x} (4 x + 6)

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