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

The integral from 0 to 1 of arccos(x) is 1

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integral from 0 to 1 of arccos(x)

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

1

Solution steps

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

Apply Integration By Parts

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

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

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

= [x arccos (x) - 1 - x^{2}]_{0}^{1}

Compute the boundaries:

1

= 1

\int_{1}^{2} \frac{x ^{2} + 2 x + 4}{x} d x

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

\int_{0}^{1} 2 y d y

\int_{- π}^{0} sin (n x) d x

\int_{0}^{3} x e^{- x^{2}} d x

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