What is the integral from 0 to pi/3 of 5sin(3x+pi) ?

The integral from 0 to pi/3 of 5sin(3x+pi) is -10/3

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integral from 0 to pi/3 of 5sin(3x+pi)

\int_{0}^{\frac{π}{3}} 5 sin (3 x + π) d x

- \frac{10}{3}

Decimal

- 3.33333 \dots

Solution steps

\int_{0}^{\frac{π}{3}} 5 sin (3 x + π) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{π}{3}} sin (3 x + π) d x

Rewrite using trig identities

= 5 \cdot \int_{0}^{\frac{π}{3}} - sin (3 x) d x

Take the constant out:

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

= 5 (- \int_{0}^{\frac{π}{3}} sin (3 x) d x)

Apply u-substitution

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

= 5 (- \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

= 5 (- \frac{1}{3} \cdot \int_{0}^{π} sin (u) d u)

Use the common integral:

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

= 5 (- \frac{1}{3} [- cos (u)]_{0}^{π})

Simplify

5 (- \frac{1}{3} [- cos (u)]_{0}^{π}) : - \frac{5}{3} [- cos (u)]_{0}^{π}

= - \frac{5}{3} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= - \frac{5}{3} \cdot 2

Simplify

= - \frac{10}{3}

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What is the integral from 0 to pi/3 of 5sin(3x+pi) ?
The integral from 0 to pi/3 of 5sin(3x+pi) is -10/3