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

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

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

\int_{0}^{\frac{π}{6}} cos (2 t) d t

\frac{3}{4}

Solution steps

\int_{0}^{\frac{π}{6}} cos (2 t) d t

Apply u-substitution

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

= \int_{0}^{\frac{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{π}{3}} cos (u) d u

Use the common integral:

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

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

Compute the boundaries:

\frac{3}{2}

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

Simplify

= \frac{3}{4}

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\int_{0}^{1} - \frac{( x - 1 ) ^{3}}{3} d x

\int_{0}^{1} 2 π (2 - x) (x - x^{2}) d x

\int_{- 4}^{0} (- x^{2} - 4 x) d x

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