What is the integral from 0 to 3 of xcos((npix)/3) ?

The integral from 0 to 3 of xcos((npix)/3) is (9(-1)^n-9)/(pi^2n^2)

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integral from 0 to 3 of xcos((npix)/3)

\int_{0}^{3} x cos (\frac{nπ x}{3}) d x

\frac{9 ( - 1 ) ^{n} - 9}{π ^{2} n ^{2}}

Solution steps

\int_{0}^{3} x cos (\frac{nπ x}{3}) d x

Apply u-substitution

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

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

Apply Integration By Parts

: [3 u sin (\frac{u}{3}) - \int 3 sin (\frac{u}{3}) d u]_{0}^{3 πn}

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

\int 3 sin (\frac{u}{3}) d u = - 9 cos (\frac{u}{3})

= \frac{1}{π ^{2} n ^{2}} [3 u sin (\frac{u}{3}) - (- 9 cos (\frac{u}{3}))]_{0}^{3 πn}

Simplify

= \frac{1}{π ^{2} n ^{2}} [3 u sin (\frac{u}{3}) + 9 cos (\frac{u}{3})]_{0}^{3 πn}

Compute the boundaries:

9 (- 1)^{n} - 9

= \frac{1}{π ^{2} n ^{2}} (9 (- 1)^{n} - 9)

Simplify

= \frac{9 ( - 1 ) ^{n} - 9}{π ^{2} n ^{2}}

\int_{0}^{3} \frac{x}{6 - x} d x

\int_{0}^{1} \frac{1}{( 3 - 2 x ) ^{2}} d x

\int_{1}^{2} (x^{2})^{2} d x

\int_{0}^{1} (3 x) d x

\int_{1}^{5} (9 x)^{3} d x

What is the integral from 0 to 3 of xcos((npix)/3) ?
The integral from 0 to 3 of xcos((npix)/3) is (9(-1)^n-9)/(pi^2n^2)