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

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

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

\int_{0}^{2 π} 3 sin (2 x) d x

0

Solution steps

\int_{0}^{2 π} 3 sin (2 x) d x

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 x) d x

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

x \to 0 lim (\frac{- ln ( 1 + 6 x )}{x})

\int \frac{x ^{3}}{1 + x ^{2} - x ^{4}} d x

\frac{d}{d x} (x^{3} sin (\frac{1}{x}))

\frac{d}{d y} (\frac{x y}{x ^{2} + y ^{2}})

f (x) = \frac{2 - 5 x}{e ^{x}}

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