What is the integral from 0 to pi of 2xsin(nx) ?

The integral from 0 to pi of 2xsin(nx) is -(2pi(-1)^n)/n

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integral from 0 to pi of 2xsin(nx)

\int_{0}^{π} 2 x sin (n x) d x

- \frac{2 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{0}^{π} 2 x sin (n x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{π} x sin (n x) d x

Apply Integration By Parts

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

= 2 [\frac{1}{n} (- x cos (n x) - 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)

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

Simplify

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

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

Compute the boundaries:

- (- 1)^{n} \frac{π}{n}

= 2 (- (- 1)^{n} \frac{π}{n})

Simplify

= - \frac{2 π ( - 1 ) ^{n}}{n}

\int_{5}^{6} x^{2} + 2 x - 16 d x

\int_{- 2}^{2} \frac{sin ( x )}{1 + x ^{2}} d x

\int_{0}^{π} ∣ cos (x + y) ∣ d x

\int_{0}^{2} \frac{- 1}{x ^{2} + 1} d x

\int_{- 2}^{2} \frac{x ^{2}}{2} d x

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