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

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

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integral from 0 to 2 of xcos((npi)/2 x)

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

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

Solution steps

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

\frac{nπ}{2} = nπ \frac{1}{2}

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

= \frac{4}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{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}^{1} \frac{1}{y + 1} d y

\int_{- 3}^{2} x^{3} d x

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

\int_{0}^{1} \frac{4}{t ^{2} + 1} d t

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

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