What is the integral from 0 to pi/2 of-sin(2t-pi/3) ?

The integral from 0 to pi/2 of-sin(2t-pi/3) is -1/2

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integral from 0 to pi/2 of-sin(2t-pi/3)

\int_{0}^{\frac{π}{2}} - sin (2 t - \frac{π}{3}) d t

- \frac{1}{2}

Decimal

- 0.5

Solution steps

\int_{0}^{\frac{π}{2}} - sin (2 t - \frac{π}{3}) d t

Take the constant out:

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

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

Apply u-substitution

: \int_{- \frac{π}{3}}^{π - \frac{π}{3}} sin (u) \frac{1}{2} d u

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

Use the common integral:

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

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

Compute the boundaries:

1

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

Simplify

= - \frac{1}{2}

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What is the integral from 0 to pi/2 of-sin(2t-pi/3) ?
The integral from 0 to pi/2 of-sin(2t-pi/3) is -1/2