What is the integral from 0 to 2 of x^2cos(npix) ?

The integral from 0 to 2 of x^2cos(npix) is (4pi^2n^2sin(2pin)-2(sin(2pin)-2pincos(2pin)))/(pi^3n^3)

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integral from 0 to 2 of x^2cos(npix)

\int_{0}^{2} x^{2} cos (nπ x) d x

\frac{4 π ^{2} n ^{2} sin ( 2 πn ) - 2 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{π ^{3} n ^{3}}

Solution steps

\int_{0}^{2} x^{2} cos (nπ x) d x

Apply u-substitution

: \int_{0}^{nπ 2} \frac{u ^{2} cos ( u )}{π ^{3} n ^{3}} d u

= \int_{0}^{nπ 2} \frac{u ^{2} cos ( u )}{π ^{3} n ^{3}} d u

Take the constant out:

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

= \frac{1}{π ^{3} n ^{3}} \cdot \int_{0}^{nπ 2} u^{2} cos (u) d u

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{nπ 2}

= \frac{1}{π ^{3} n ^{3}} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{nπ 2}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{π ^{3} n ^{3}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{nπ 2}

Compute the boundaries:

4 π^{2} n^{2} sin (2 πn) - 2 (sin (2 πn) - 2 πn cos (2 πn))

= \frac{1}{π ^{3} n ^{3}} (4 π^{2} n^{2} sin (2 πn) - 2 (sin (2 πn) - 2 πn cos (2 πn)))

Simplify

= \frac{4 π ^{2} n ^{2} sin ( 2 πn ) - 2 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{π ^{3} n ^{3}}

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What is the integral from 0 to 2 of x^2cos(npix) ?
The integral from 0 to 2 of x^2cos(npix) is (4pi^2n^2sin(2pin)-2(sin(2pin)-2pincos(2pin)))/(pi^3n^3)