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

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

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

\int_{0}^{2 π} 3 sin (2 t) d t

0

Solution steps

\int_{0}^{2 π} 3 sin (2 t) d t

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{4 π} sin (u) \frac{1}{2} d u

= 3 \cdot \int_{0}^{4 π} 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

= 3 \cdot \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π}

Simplify

3 \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π} : \frac{3}{2} [- cos (u)]_{0}^{4 π}

= \frac{3}{2} [- cos (u)]_{0}^{4 π}

Compute the boundaries:

0

= \frac{3}{2} \cdot 0

Simplify

= 0

\frac{d}{d x} (ln (\frac{8 x}{x + 5}))

x \to 5 lim (\frac{1}{( x - 5 ) ^{3}})

\frac{\partial}{\partial x} (\frac{e ^{x}}{y + x ^{8}})

\frac{d}{d x} (\frac{x + y}{x y})

y^{^{'}} = 1 - \frac{y}{20}

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