What is the integral from 0 to (5pi)/6 of cos(2t) ?

The integral from 0 to (5pi)/6 of cos(2t) is -(sqrt(3))/4

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integral from 0 to (5pi)/6 of cos(2t)

\int_{0}^{\frac{5 π}{6}} cos (2 t) d t

- \frac{3}{4}

Solution steps

\int_{0}^{\frac{5 π}{6}} cos (2 t) d t

Apply u-substitution

: \int_{0}^{\frac{5 π}{3}} cos (u) \frac{1}{2} d u

= \int_{0}^{\frac{5 π}{3}} cos (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_{0}^{\frac{5 π}{3}} cos (u) d u

Use the common integral:

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

= \frac{1}{2} [sin (u)]_{0}^{\frac{5 π}{3}}

Compute the boundaries:

- \frac{3}{2}

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

Simplify

= - \frac{3}{4}

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What is the integral from 0 to (5pi)/6 of cos(2t) ?
The integral from 0 to (5pi)/6 of cos(2t) is -(sqrt(3))/4