What is the integral from 0 to 2 of xcos(pi/2 yx) ?

The integral from 0 to 2 of xcos(pi/2 yx) is (4(piysin(piy)+cos(piy)-1))/(pi^2y^2)

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integral from 0 to 2 of xcos(pi/2 yx)

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

\frac{4 ( π y sin ( π y ) + cos ( π y ) - 1 )}{π ^{2} y ^{2}}

Solution steps

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

Apply u-substitution

: \int_{0}^{π y} \frac{4 u cos ( u )}{π ^{2} y ^{2}} d u

= \int_{0}^{π y} \frac{4 u cos ( u )}{π ^{2} y ^{2}} d u

Take the constant out:

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

= \frac{4}{π ^{2} y ^{2}} \cdot \int_{0}^{π y} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{π y}

= \frac{4}{π ^{2} y ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{π y}

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

= \frac{4}{π ^{2} y ^{2}} [u sin (u) - (- cos (u))]_{0}^{π y}

Simplify

= \frac{4}{π ^{2} y ^{2}} [u sin (u) + cos (u)]_{0}^{π y}

Compute the boundaries:

π y sin (π y) + cos (π y) - 1

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

Simplify

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

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What is the integral from 0 to 2 of xcos(pi/2 yx) ?
The integral from 0 to 2 of xcos(pi/2 yx) is (4(piysin(piy)+cos(piy)-1))/(pi^2y^2)