What is the integral from 0 to pi/6 of sqrt(1-cos(6x)) ?

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

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integral from 0 to pi/6 of sqrt(1-cos(6x))

\int_{0}^{\frac{π}{6}} 1 - cos (6 x) d x

\frac{2}{3}

Decimal Notation

0.47140 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 1 - cos (6 x) d x

Apply Integral Substitution

= \int_{0}^{\frac{π}{3}} \frac{1}{2} 1 - cos (3 u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} 1 - cos (3 u) d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int_{0}^{\frac{2 π}{3}} \frac{1}{2} 1 - cos (\frac{3 v}{2}) d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{2 π}{3}} 1 - cos (\frac{3 v}{2}) d v

Apply u-substitution

: \int_{0}^{π} 1 - cos (w) \frac{2}{3} d w

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 - cos (w) \frac{2}{3} d w

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \int_{0}^{π} 1 - cos (w) d w

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \int_{0}^{π} 2 sin^{2} (\frac{w}{2}) d w

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{π} sin^{2} (\frac{w}{2}) d w

Apply exponent rule:

a^{2} = ∣ a ∣

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{π} sin (\frac{w}{2}) d w

Eliminate Absolutes

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{π} sin (\frac{w}{2}) d w

Apply u-substitution

: \int_{0}^{\frac{π}{2}} sin (t) \cdot 2 d t

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot \int_{0}^{\frac{π}{2}} sin (t) \cdot 2 d t

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot 2 \cdot \int_{0}^{\frac{π}{2}} sin (t) d t

Use the common integral:

\int sin (t) d t = - cos (t)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot 2 [- cos (t)]_{0}^{\frac{π}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} 2 \cdot 2 [- cos (t)]_{0}^{\frac{π}{2}} : \frac{2}{3} [- cos (t)]_{0}^{\frac{π}{2}}

= \frac{2}{3} [- cos (t)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{2}{3} \cdot 1

Simplify

= \frac{2}{3}

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