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

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

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

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

\frac{π}{6} - \frac{3}{2} + 1

Decimal Notation

0.65757 \dots

Solution steps

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

Simplify

= \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

e^{z}

x \frac{d y}{d x} - y = x^{2} sin (x)

y = \frac{x}{5 + x}

\int_{- 1}^{1} 7^{x} d x

x \to - 1 lim (\frac{x}{x + 1})

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