What is the integral from 1 to 3 of 12sin(6x-5) ?

The integral from 1 to 3 of 12sin(6x-5) is 2(-cos(13)+cos(1))

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integral from 1 to 3 of 12sin(6x-5)

\int_{1}^{3} 12 sin (6 x - 5) d x

2 (- cos (13) + cos (1))

Decimal Notation

- 0.73428 \dots

Solution steps

\int_{1}^{3} 12 sin (6 x - 5) d x

Take the constant out:

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

= 12 \cdot \int_{1}^{3} sin (6 x - 5) d x

Apply u-substitution

: \int_{1}^{13} sin (u) \frac{1}{6} d u

= 12 \cdot \int_{1}^{13} sin (u) \frac{1}{6} d u

Take the constant out:

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

= 12 \cdot \frac{1}{6} \cdot \int_{1}^{13} sin (u) d u

Use the common integral:

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

= 12 \cdot \frac{1}{6} [- cos (u)]_{1}^{13}

Simplify

12 \cdot \frac{1}{6} [- cos (u)]_{1}^{13} : 2 [- cos (u)]_{1}^{13}

= 2 [- cos (u)]_{1}^{13}

Compute the boundaries:

- cos (13) + cos (1)

= 2 (- cos (13) + cos (1))

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What is the integral from 1 to 3 of 12sin(6x-5) ?
The integral from 1 to 3 of 12sin(6x-5) is 2(-cos(13)+cos(1))