What is the integral from-pi to pi of x^4cos(nx) ?

The integral from-pi to pi of x^4cos(nx) is -(48pi(-1)^n)/(n^4)+(8pi^3(-1)^n)/(n^2)

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integral from-pi to pi of x^4cos(nx)

\int_{- π}^{π} x^{4} cos (n x) d x

- \frac{48 π ( - 1 ) ^{n}}{n ^{4}} + \frac{8 π ^{3} ( - 1 ) ^{n}}{n ^{2}}

Solution steps

\int_{- π}^{π} x^{4} cos (n x) d x

Apply Integration By Parts

: [\frac{1}{n} (x^{4} sin (n x) - n \cdot \int \frac{4}{n} x^{3} sin (n x) d x)]_{- π}^{π}

= [\frac{1}{n} (x^{4} sin (n x) - n \cdot \int \frac{4}{n} x^{3} sin (n x) d x)]_{- π}^{π}

\int \frac{4}{n} x^{3} sin (n x) d x = \frac{4}{n} (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x))))

= [\frac{1}{n} (x^{4} sin (n x) - n \frac{4}{n} (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{- π}^{π}

Simplify

[\frac{1}{n} (x^{4} sin (n x) - n \frac{4}{n} (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{- π}^{π} : [\frac{1}{n} (x^{4} sin (n x) - 4 (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{- π}^{π}

= [\frac{1}{n} (x^{4} sin (n x) - 4 (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{- π}^{π}

Compute the boundaries:

- \frac{48 π ( - 1 ) ^{n}}{n ^{4}} + (- 1)^{n} \frac{8 π ^{3}}{n ^{2}}

= - \frac{48 π ( - 1 ) ^{n}}{n ^{4}} + (- 1)^{n} \frac{8 π ^{3}}{n ^{2}}

Simplify

= - \frac{48 π ( - 1 ) ^{n}}{n ^{4}} + \frac{8 π ^{3} ( - 1 ) ^{n}}{n ^{2}}

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What is the integral from-pi to pi of x^4cos(nx) ?
The integral from-pi to pi of x^4cos(nx) is -(48pi(-1)^n)/(n^4)+(8pi^3(-1)^n)/(n^2)