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

The integral from 0 to pi/6 of sin(3x) is 1/3

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integral from 0 to pi/6 of sin(3x)

\int_{0}^{\frac{π}{6}} sin (3 x) d x

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

\int_{0}^{\frac{π}{6}} sin (3 x) d x

Apply u-substitution

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

= \int_{0}^{\frac{π}{2}} sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{\frac{π}{2}} sin (u) d u

Use the common integral:

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

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

Compute the boundaries:

1

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

Simplify

= \frac{1}{3}

\frac{d}{d x} (\frac{- x ^{3} + 10}{4})

\int e^{2 s i n (3 x)} cos (3 x) d x

\frac{\partial}{\partial x} (2 y^{\frac{1}{2}} x^{\frac{1}{2}})

f (x) = \frac{10}{x}

x \to 2 lim (\frac{x ^{2}}{x ^{2} - 4})

What is the integral from 0 to pi/6 of sin(3x) ?
The integral from 0 to pi/6 of sin(3x) is 1/3