What is the integral from 0 to 1/2 of 5cos(-1(x)) ?

The integral from 0 to 1/2 of 5cos(-1(x)) is 5sin(1/2)

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integral from 0 to 1/2 of 5cos(-1(x))

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

5 sin (\frac{1}{2})

Decimal Notation

2.39712 \dots

Solution steps

\int_{0}^{\frac{1}{2}} 5 cos (- 1 \cdot x) 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{1}{2}} cos (- 1 \cdot x) d x

Simplify

cos (- 1 \cdot x) : cos (x)

= 5 \cdot \int_{0}^{\frac{1}{2}} cos (x) d x

Use the common integral:

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

= 5 [sin (x)]_{0}^{\frac{1}{2}}

Compute the boundaries:

sin (\frac{1}{2})

= 5 sin (\frac{1}{2})

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

\frac{\partial}{\partial x} (e^{- \frac{π}{2}})

\int 20 t e^{t} d t

f (x) = 4 x^{2} + 7, (- 3, 43)

(- 4, - 6), (4, - 2)

What is the integral from 0 to 1/2 of 5cos(-1(x)) ?
The integral from 0 to 1/2 of 5cos(-1(x)) is 5sin(1/2)