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

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

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

\int_{0}^{\frac{π}{6}} cos (2 x) d x

\frac{3}{4}

Decimal

0.43301 \dots

Solution steps

\int_{0}^{\frac{π}{6}} cos (2 x) d x

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}

\int_{- \frac{π}{3}}^{\frac{π}{3}} tan (x) d x

\int_{0}^{8} \frac{x}{49 + 4 x} d x

\int_{- t}^{t^{2}} e^{- x} d x

\int_{- 1}^{x^{2}} (1 + t^{3})^{4} d t

\int_{0}^{4} π (4 - z) d z

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