What is the integral from 3 to 7 of 4cos(2x+3) ?

The integral from 3 to 7 of 4cos(2x+3) is 2(sin(17)-sin(9))

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integral from 3 to 7 of 4cos(2x+3)

\int_{3}^{7} 4 cos (2 x + 3) d x

2 (sin (17) - sin (9))

Decimal Notation

- 2.74703 \dots

Solution steps

\int_{3}^{7} 4 cos (2 x + 3) d x

Take the constant out:

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

= 4 \cdot \int_{3}^{7} cos (2 x + 3) d x

Apply u-substitution

: \int_{9}^{17} cos (u) \frac{1}{2} d u

= 4 \cdot \int_{9}^{17} cos (u) \frac{1}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{9}^{17} cos (u) d u

Use the common integral:

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

= 4 \cdot \frac{1}{2} [sin (u)]_{9}^{17}

Simplify

4 \cdot \frac{1}{2} [sin (u)]_{9}^{17} : 2 [sin (u)]_{9}^{17}

= 2 [sin (u)]_{9}^{17}

Compute the boundaries:

sin (17) - sin (9)

= 2 (sin (17) - sin (9))

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What is the integral from 3 to 7 of 4cos(2x+3) ?
The integral from 3 to 7 of 4cos(2x+3) is 2(sin(17)-sin(9))