What is the integral from-pi to pi of (1+x)cos(nx) ?

The integral from-pi to pi of (1+x)cos(nx) is 0

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integral from-pi to pi of (1+x)cos(nx)

\int_{- π}^{π} (1 + x) cos (n x) d x

0

Solution steps

\int_{- π}^{π} (1 + x) cos (n x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

0

= 0

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What is the integral from-pi to pi of (1+x)cos(nx) ?
The integral from-pi to pi of (1+x)cos(nx) is 0