What is the integral from pi/(12) to pi/9 of sin(3x) ?

The integral from pi/(12) to pi/9 of sin(3x) is (-1+sqrt(2))/6

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integral from pi/(12) to pi/9 of sin(3x)

\int_{\frac{π}{12}}^{\frac{π}{9}} sin (3 x) d x

\frac{- 1 + 2}{6}

Decimal

0.06903 \dots

Solution steps

\int_{\frac{π}{12}}^{\frac{π}{9}} sin (3 x) d x

Apply u-substitution

: \int_{\frac{π}{4}}^{\frac{π}{3}} sin (u) \frac{1}{3} d u

= \int_{\frac{π}{4}}^{\frac{π}{3}} sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} sin (u) d u

Use the common integral:

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

= \frac{1}{3} [- cos (u)]_{\frac{π}{4}}^{\frac{π}{3}}

Compute the boundaries:

- \frac{1}{2} + \frac{1}{2}

= \frac{1}{3} (- \frac{1}{2} + \frac{1}{2})

Simplify

= \frac{- 1 + 2}{6}

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What is the integral from pi/(12) to pi/9 of sin(3x) ?
The integral from pi/(12) to pi/9 of sin(3x) is (-1+sqrt(2))/6