What is the integral from 0 to 1/2 of 6arccos(x) ?

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

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integral from 0 to 1/2 of 6arccos(x)

\int_{0}^{\frac{1}{2}} 6 arccos (x) d x

π - 33 + 6

Decimal Notation

3.94544 \dots

Solution steps

\int_{0}^{\frac{1}{2}} 6 arccos (x) d x

Take the constant out:

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

= 6 \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}}

= 6 [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}

= 6 [x arccos (x) - 1 - x^{2}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{π}{6} - \frac{3}{2} + 1

= 6 (\frac{π}{6} - \frac{3}{2} + 1)

Simplify

= π - 33 + 6

x \to \infty lim (5 5^{x})

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

y = (1 + 3 x)^{10}, (0, 1)

x \to 0 lim (2 e^{5 x})

\int_{0}^{7} - 30 x^{2} + 340 x d x

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