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

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

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

\int_{0}^{\frac{1}{2}} cos (π x) d x

\frac{1}{π}

Decimal

0.31830 \dots

Solution steps

\int_{0}^{\frac{1}{2}} cos (π x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{π} d u

= \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{π} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{π} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{1}{π} \cdot 1

Simplify

= \frac{1}{π}

(2 x y + 3 y^{2}) d x - (x^{2} + 2 x y) d y = 0

f (x) = 34 x - 16 x^{2}, at x = 1

10 e^{10 x}

\frac{\partial}{\partial m} (\frac{G ( m , r ) m}{r ^{2}})

x \to 0 - lim (x \cdot e^{\frac{1}{x}})

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