What is the integral from 0 to 1.2 of 9cos(1.9*x) ?

The integral from 0 to 1.2 of 9cos(1.9*x) is 3.59469…

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integral from 0 to 1.2 of 9cos(1.9*x)

\int_{0}^{1.2} 9 cos (1.9 \cdot x) d x

3.59469 \dots

Solution steps

\int_{0}^{1.2} 9 cos (1.9 x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{1.2} cos (1.9 x) d x

Apply u-substitution

: \int_{0}^{2.28} cos (u) \frac{1}{1.9} d u

= 9 \cdot \int_{0}^{2.28} cos (u) \frac{1}{1.9} d u

Take the constant out:

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

= 9 \cdot \frac{1}{1.9} \cdot \int_{0}^{2.28} cos (u) d u

Simplify

= 9 \cdot 0.52631 \dots \cdot \int_{0}^{2.28} cos (u) d u

Use the common integral:

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

= 9 \cdot 0.52631 \dots [sin (u)]_{0}^{2.28}

Simplify

= 4.73684 \dots [sin (u)]_{0}^{2.28}

Compute the boundaries:

0.75888 \dots

= 4.73684 \dots \cdot 0.75888 \dots

Simplify

= 3.59469 \dots

\int sin (3 x) \cdot e^{2 x} d x

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\frac{d}{d x} (2 x^{3} y^{2})

\frac{\partial}{\partial x} (cos (x))

y = x (x - 2)^{3}

What is the integral from 0 to 1.2 of 9cos(1.9*x) ?
The integral from 0 to 1.2 of 9cos(1.9*x) is 3.59469…