What is the integral from 0 to pi/6 of-xcos(x+pi/3) ?

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

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integral from 0 to pi/6 of-xcos(x+pi/3)

\int_{0}^{\frac{π}{6}} - x cos (x + \frac{π}{3}) d x

- \frac{π}{6} + \frac{1}{2}

Decimal

- 0.02359 \dots

Solution steps

\int_{0}^{\frac{π}{6}} - x cos (x + \frac{π}{3}) d x

Take the constant out:

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

= - \int_{0}^{\frac{π}{6}} x cos (x + \frac{π}{3}) d x

Apply Integration By Parts

: [x sin (x + \frac{π}{3}) - \int sin (x + \frac{π}{3}) d x]_{0}^{\frac{π}{6}}

= - [x sin (x + \frac{π}{3}) - \int sin (x + \frac{π}{3}) d x]_{0}^{\frac{π}{6}}

\int sin (x + \frac{π}{3}) d x = - cos (x + \frac{π}{3})

= - [x sin (x + \frac{π}{3}) - (- cos (x + \frac{π}{3}))]_{0}^{\frac{π}{6}}

Simplify

= - [x sin (x + \frac{π}{3}) + cos (x + \frac{π}{3})]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6} - \frac{1}{2}

= - (\frac{π}{6} - \frac{1}{2})

Simplify

= - \frac{π}{6} + \frac{1}{2}

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