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

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

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

\int_{0}^{\frac{π}{3}} (sin (3 x)) d x

\frac{2}{3}

Solution steps

\int_{0}^{\frac{π}{3}} sin (3 x) d x

Apply u-substitution

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

= \int_{0}^{π} 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}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{3} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

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

Simplify

= \frac{2}{3}

\int_{0}^{\frac{π}{3}} cos (x) - 0.5 d x

\int_{0}^{1} 12 x^{2} (1 - x) d x

\int_{0}^{3} x^{2} - 4 x + 3 d x

\int_{5}^{25} \frac{1}{x ( ln ( x ) ) ^{9}} d x

\int_{0}^{5} \frac{1}{1 - x ^{2}} d x

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