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

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

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

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

- \frac{π}{3} + 3 - 2

Decimal Notation

- 1.31514 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \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}}

= - 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}

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

Compute the boundaries:

\frac{π}{6} - \frac{3}{2} + 1

= - 2 (\frac{π}{6} - \frac{3}{2} + 1)

Simplify

= - \frac{π}{3} + 3 - 2

\int_{0}^{\infty} 9 e^{- 9 x} d x

\frac{\partial}{\partial x} (1 - \frac{3}{y} + x)

\int_{1}^{8} 1 + 3 x d x

\frac{d y}{d x} = ka e^{- k x} - b y

\frac{s}{( s ^{2} - 1 ^{2} ) ^{2}}

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