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}}

\int_{0}^{4} (1 - e^{- x}) d x

\int_{0}^{1} x ln (x) d x

\int_{1}^{2} \frac{1}{x ^{2} - 1} d x

\int_{1}^{e} \frac{1}{x ( 1 + ln ( x ) )} d x

\int_{- 1}^{1} 2 x^{3} 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)