What is the integral from-1 to 1 of (arccos(x))^2 ?

The integral from-1 to 1 of (arccos(x))^2 is pi^2-4

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integral from-1 to 1 of (arccos(x))^2

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

π^{2} - 4

Decimal

5.86960 \dots

Solution steps

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

Simplify

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

Apply Integration By Parts

: [x arccos^{2} (x) - \int - \frac{2 x arccos ( x )}{1 - x ^{2}} d x]_{- 1}^{1}

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

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

= [x arccos^{2} (x) - 2 (arccos (x) 1 - x^{2} + x)]_{- 1}^{1}

Compute the boundaries:

π^{2} - 4

= π^{2} - 4

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What is the integral from-1 to 1 of (arccos(x))^2 ?
The integral from-1 to 1 of (arccos(x))^2 is pi^2-4