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

The integral from-1/2 to 1/2 of 2arccos(2x) is pi

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

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

π

Decimal

3.14159 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: [x arccos (2 x) - \int - \frac{2 x}{1 - 4 x ^{2}} d x]_{- \frac{1}{2}}^{\frac{1}{2}}

= 2 [x arccos (2 x) - \int - \frac{2 x}{1 - 4 x ^{2}} d x]_{- \frac{1}{2}}^{\frac{1}{2}}

\int - \frac{2 x}{1 - 4 x ^{2}} d x = \frac{1}{2} 1 - 4 x^{2}

= 2 [x arccos (2 x) - \frac{1}{2} 1 - 4 x^{2}]_{- \frac{1}{2}}^{\frac{1}{2}}

Simplify

2 [x arccos (2 x) - \frac{1}{2} 1 - 4 x^{2}]_{- \frac{1}{2}}^{\frac{1}{2}} : 2 [\frac{1}{2} (2 x arccos (2 x) - 1 - 4 x^{2})]_{- \frac{1}{2}}^{\frac{1}{2}}

= 2 [\frac{1}{2} (2 x arccos (2 x) - 1 - 4 x^{2})]_{- \frac{1}{2}}^{\frac{1}{2}}

Compute the boundaries:

\frac{π}{2}

= 2 \cdot \frac{π}{2}

Simplify

= π

\frac{d ^{2}}{d x ^{2}} (x^{3} - \frac{3}{x})

\frac{d}{d x} (arccot (x^{3}))

\int 8 x + 1 - 9 e^{x} d x

x \to 0 lim ((x + 1) \frac{3}{x})

f (x) = x^{3} - 3 x + 2, (4, 54)

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