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

The integral from 0 to 1 of xcos((npix)/2) is (2(pinsin((pin)/2)+2cos((pin)/2)-2))/(pi^2n^2)

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

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

\frac{2 ( πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 )}{π ^{2} n ^{2}}

Solution steps

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

Apply u-substitution

: \int_{0}^{πn} \frac{u cos ( \frac{u}{2} )}{π ^{2} n ^{2}} d u

= \int_{0}^{πn} \frac{u cos ( \frac{u}{2} )}{π ^{2} n ^{2}} d u

Take the constant out:

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

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{0}^{πn} u cos (\frac{u}{2}) d u

Apply u-substitution

: \int_{0}^{\frac{πn}{2}} 4 v cos (v) d v

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{0}^{\frac{πn}{2}} 4 v cos (v) d v

Take the constant out:

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

= \frac{1}{π ^{2} n ^{2}} \cdot 4 \cdot \int_{0}^{\frac{πn}{2}} v cos (v) d v

Apply Integration By Parts

: [v sin (v) - \int sin (v) d v]_{0}^{\frac{πn}{2}}

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - \int sin (v) d v]_{0}^{\frac{πn}{2}}

\int sin (v) d v = - cos (v)

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{0}^{\frac{πn}{2}}

Simplify

\frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{0}^{\frac{πn}{2}} : \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{0}^{\frac{πn}{2}}

= \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{0}^{\frac{πn}{2}}

Compute the boundaries:

\frac{πn}{2} sin (\frac{πn}{2}) + cos (\frac{πn}{2}) - 1

= \frac{4}{π ^{2} n ^{2}} (\frac{πn}{2} sin (\frac{πn}{2}) + cos (\frac{πn}{2}) - 1)

Simplify

= \frac{2 ( πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 )}{π ^{2} n ^{2}}

ln (\frac{1}{2 x + 3})

\frac{1}{( 5 \cdot s + 1 )}

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

\frac{d}{d x} (\frac{a x}{x ^{2} + a ^{2}})

\int (x^{4} + 2 x - 16) d x

What is the integral from 0 to 1 of xcos((npix)/2) ?
The integral from 0 to 1 of xcos((npix)/2) is (2(pinsin((pin)/2)+2cos((pin)/2)-2))/(pi^2n^2)