What is the integral from 0 to 1 of 8cos((pix)/2) ?

The integral from 0 to 1 of 8cos((pix)/2) is (16)/pi

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integral from 0 to 1 of 8cos((pix)/2)

\int_{0}^{1} 8 cos (\frac{π x}{2}) d x

\frac{16}{π}

Decimal

5.09295 \dots

Solution steps

\int_{0}^{1} 8 cos (\frac{π x}{2}) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} cos (\frac{π x}{2}) d x

Apply u-substitution

= 8 \cdot \int_{0}^{\frac{π}{2}} cos (u) \frac{2}{π} d u

Take the constant out:

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

= 8 \cdot \frac{2}{π} \cdot \int_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

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

= 8 \cdot \frac{2}{π} [sin (u)]_{0}^{\frac{π}{2}}

Simplify

8 \cdot \frac{2}{π} [sin (u)]_{0}^{\frac{π}{2}} : \frac{16}{π} [sin (u)]_{0}^{\frac{π}{2}}

= \frac{16}{π} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{16}{π} \cdot 1

Simplify

= \frac{16}{π}

What is the integral from 0 to 1 of 8cos((pix)/2) ?
The integral from 0 to 1 of 8cos((pix)/2) is (16)/pi