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

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

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integral from 0 to pi of (x+1)sin(nx)

\int_{0}^{π} (x + 1) sin (n x) d x

\frac{- ( π + 1 ) ( - 1 ) ^{n} + 1}{n}

Solution steps

\int_{0}^{π} (x + 1) sin (n x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- (- 1)^{n} \frac{π + 1}{n} + \frac{1}{n}

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

Simplify

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

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