What is the integral from 0 to 1/2 of-7arccos(x) ?

The integral from 0 to 1/2 of-7arccos(x) is -7(pi/6-(sqrt(3))/2+1)

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integral from 0 to 1/2 of-7arccos(x)

\int_{0}^{\frac{1}{2}} - 7 arccos (x) d x

- 7 (\frac{π}{6} - \frac{3}{2} + 1)

Decimal

- 4.60301 \dots

Solution steps

\int_{0}^{\frac{1}{2}} - 7 arccos (x) d x

Take the constant out:

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

= - 7 \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}}

= - 7 [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}

= - 7 [x arccos (x) - 1 - x^{2}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{π}{6} - \frac{3}{2} + 1

= - 7 (\frac{π}{6} - \frac{3}{2} + 1)

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What is the integral from 0 to 1/2 of-7arccos(x) ?
The integral from 0 to 1/2 of-7arccos(x) is -7(pi/6-(sqrt(3))/2+1)