What is the integral from 0 to 1 of (arccos(x))^2 ?

The integral from 0 to 1 of (arccos(x))^2 is -2+pi

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integral from 0 to 1 of (arccos(x))^2

\int_{0}^{1} (arccos (x))^{2} d x

- 2 + π

Decimal

1.14159 \dots

Solution steps

\int_{0}^{1} (arccos (x))^{2} d x

Simplify

= \int_{0}^{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]_{0}^{1}

= [x arccos^{2} (x) - \int - \frac{2 x arccos ( x )}{1 - x ^{2}} d x]_{0}^{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)]_{0}^{1}

Compute the boundaries:

- 2 + π

= - 2 + π

\int_{- 2}^{4} (2 x - 1) d x

\int_{0}^{2 π} x (cos (x))^{2} d x

\int_{- 2}^{3} (4 x^{5} - 5) d x

\int_{3}^{5} \frac{6}{x ^{2}} d x

\int_{5}^{10} - 0.6 x^{2} + 4 x + 4 d x

What is the integral from 0 to 1 of (arccos(x))^2 ?
The integral from 0 to 1 of (arccos(x))^2 is -2+pi