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

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

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integral from-pi to 0 of 2xsin(nx)

\int_{- π}^{0} 2 x sin (n x) d x

- \frac{2 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{0} 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_{- π}^{0} x sin (n x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

- (- 1)^{n} \frac{π}{n}

= 2 (- (- 1)^{n} \frac{π}{n})

Simplify

= - \frac{2 π ( - 1 ) ^{n}}{n}

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\int_{- π}^{π} x sin (x) cos (n x) d x

\int_{1}^{8} 8 x^{\frac{1}{3}} d x

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

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