What is the integral from 0 to pi/2 of cos(kx) ?

The integral from 0 to pi/2 of cos(kx) is (sin((pik)/2))/(k)

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integral from 0 to pi/2 of cos(kx)

\int_{0}^{\frac{π}{2}} cos (k x) d x

\frac{sin ( \frac{πk}{2} )}{k}

Solution steps

\int_{0}^{\frac{π}{2}} cos (k x) d x

Apply u-substitution

: \int_{0}^{\frac{πk}{2}} \frac{cos ( u )}{k} d u

= \int_{0}^{\frac{πk}{2}} \frac{cos ( u )}{k} d u

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

sin (\frac{πk}{2})

= \frac{1}{k} sin (\frac{πk}{2})

Simplify

= \frac{sin ( \frac{πk}{2} )}{k}

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What is the integral from 0 to pi/2 of cos(kx) ?
The integral from 0 to pi/2 of cos(kx) is (sin((pik)/2))/(k)