What is the integral from 0 to pi of (10+x)cos(nx) ?

The integral from 0 to pi of (10+x)cos(nx) is ((-1)^n-1)/(n^2)

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integral from 0 to pi of (10+x)cos(nx)

\int_{0}^{π} (10 + x) cos (n x) d x

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

Solution steps

\int_{0}^{π} (10 + x) cos (n x) d x

Apply Integration By Parts

: [\frac{1}{n} (sin (n x) (10 + x) - n \cdot \int \frac{1}{n} sin (n x) d x)]_{0}^{π}

= [\frac{1}{n} (sin (n x) (10 + x) - n \cdot \int \frac{1}{n} sin (n x) d x)]_{0}^{π}

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

= [\frac{1}{n} (sin (n x) (10 + x) - n (- \frac{1}{n ^{2}} cos (n x)))]_{0}^{π}

Simplify

[\frac{1}{n} (sin (n x) (10 + x) - n (- \frac{1}{n ^{2}} cos (n x)))]_{0}^{π} : [\frac{1}{n} (sin (n x) (10 + x) + \frac{1}{n} cos (n x))]_{0}^{π}

= [\frac{1}{n} (sin (n x) (10 + x) + \frac{1}{n} cos (n x))]_{0}^{π}

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 0 to pi of (10+x)cos(nx) ?
The integral from 0 to pi of (10+x)cos(nx) is ((-1)^n-1)/(n^2)