What is the integral from 1/2 to 1/3 of cos(pix) ?

The integral from 1/2 to 1/3 of cos(pix) is (i(sin((ip)/2)-sin((ip)/3)))/(p)

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integral from 1/2 to 1/3 of cos(pix)

\int_{\frac{1}{2}}^{\frac{1}{3}} cos (p i x) d x

\frac{i ( sin ( \frac{i p}{2} ) - sin ( \frac{i p}{3} ) )}{p}

Solution steps

\int_{\frac{1}{2}}^{\frac{1}{3}} cos (p i x) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{3}}^{\frac{1}{2}} cos (p i x) d x

Apply u-substitution

: \int_{\frac{i p}{3}}^{\frac{i p}{2}} - \frac{i cos ( u )}{p} d u

= - \int_{\frac{i p}{3}}^{\frac{i p}{2}} - \frac{i cos ( u )}{p} d u

Take the constant out:

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

= - (- \frac{i}{p} \cdot \int_{\frac{i p}{3}}^{\frac{i p}{2}} cos (u) d u)

Use the common integral:

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

= - (- \frac{i}{p} [sin (u)]_{\frac{i p}{3}}^{\frac{i p}{2}})

Simplify

= \frac{i}{p} [sin (u)]_{\frac{i p}{3}}^{\frac{i p}{2}}

Compute the boundaries:

sin (\frac{i p}{2}) - sin (\frac{i p}{3})

= \frac{i}{p} (sin (\frac{i p}{2}) - sin (\frac{i p}{3}))

Simplify

= \frac{i ( sin ( \frac{i p}{2} ) - sin ( \frac{i p}{3} ) )}{p}

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What is the integral from 1/2 to 1/3 of cos(pix) ?
The integral from 1/2 to 1/3 of cos(pix) is (i(sin((ip)/2)-sin((ip)/3)))/(p)