What is the integral from-1 to-1/2 of-cos(kpix) ?

The integral from-1 to-1/2 of-cos(kpix) is (sin((pik)/2))/(pik)

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integral from-1 to-1/2 of-cos(kpix)

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

\frac{sin ( \frac{πk}{2} )}{πk}

Solution steps

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

Take the constant out:

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

= - \int_{- 1}^{- \frac{1}{2}} cos (kπ x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

- sin (\frac{πk}{2})

= - \frac{1}{πk} (- sin (\frac{πk}{2}))

Simplify

= \frac{sin ( \frac{πk}{2} )}{πk}

\int sin (x) - 1 d x

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\frac{d ^{2} y}{d x ^{2}} = y

\int - cos (x) ln (x) d x

\int \frac{x}{( 36 - x ^{2} ) ^{3}} d x

What is the integral from-1 to-1/2 of-cos(kpix) ?
The integral from-1 to-1/2 of-cos(kpix) is (sin((pik)/2))/(pik)