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

The integral from 0 to 1/2 of arccos(x) is pi/6-(sqrt(3))/2+1

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

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

\frac{π}{6} - \frac{3}{2} + 1

Decimal

0.65757 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

\frac{π}{6} - \frac{3}{2} + 1

= \frac{π}{6} - \frac{3}{2} + 1

(x, y) \to (0, 0) lim (\frac{y}{x})

f (x) = (x^{2} - 48)^{4}

\frac{d}{d x} (6 x^{\frac{1}{3}})

\int 11 tan^{3} (θ) d θ

\int \frac{( x ^{3} )}{x ^{2} + 1} d x

What is the integral from 0 to 1/2 of arccos(x) ?
The integral from 0 to 1/2 of arccos(x) is pi/6-(sqrt(3))/2+1