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

The integral from 0 to 1/2 of xcos(pix) is 1/(2pi)-1/(pi^2)

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

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

\frac{1}{2 π} - \frac{1}{π ^{2}}

Decimal Notation

0.05783 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{π} (x sin (π x) - π \cdot \int \frac{1}{π} sin (π x) d x)]_{0}^{\frac{1}{2}}

= [\frac{1}{π} (x sin (π x) - π \cdot \int \frac{1}{π} sin (π x) d x)]_{0}^{\frac{1}{2}}

\int \frac{1}{π} sin (π x) d x = - \frac{1}{π ^{2}} cos (π x)

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

Simplify

[\frac{1}{π} (x sin (π x) - π (- \frac{1}{π ^{2}} cos (π x)))]_{0}^{\frac{1}{2}} : [\frac{1}{π} (x sin (π x) + \frac{1}{π} cos (π x))]_{0}^{\frac{1}{2}}

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

Compute the boundaries:

\frac{1}{2 π} - \frac{1}{π ^{2}}

= \frac{1}{2 π} - \frac{1}{π ^{2}}

\frac{\partial}{\partial x} (cos^{6} (x^{9} y^{8}))

\frac{\partial}{\partial x} (ln (x^{4} + y^{9}))

\frac{\partial}{\partial x} (x \cdot e^{- y})

\frac{d y}{d x} = 1

\int \frac{( x + 1 ) ^{4}}{2 x} d x

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