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

The integral from 0 to 2 of xcos((npix)/2) is (4((-1)^n-1))/(pi^2n^2)

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

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

\frac{4 ( ( - 1 ) ^{n} - 1 )}{π ^{2} n ^{2}}

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int_{0}^{πn} 4 v cos (v) d v

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

(- 1)^{n} - 1

= \frac{4}{π ^{2} n ^{2}} ((- 1)^{n} - 1)

Simplify

= \frac{4 ( ( - 1 ) ^{n} - 1 )}{π ^{2} n ^{2}}

\int_{0}^{5} \frac{x ^{3}}{x ^{2} + 1} d x

\int_{2}^{3} 4 x e^{x^{2} - 5} d x

\int_{2}^{4} \frac{x}{x - 2} d x

\int_{1}^{2} 4 x^{3} d x

\int_{1}^{5} x 5 x^{2} - 4 d x

What is the integral from 0 to 2 of xcos((npix)/2) ?
The integral from 0 to 2 of xcos((npix)/2) is (4((-1)^n-1))/(pi^2n^2)