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

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

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

\int_{- π}^{0} - x sin (n x) d x

\frac{π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{0} - x sin (n x) d x

Take the constant out:

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

= - \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}

= - [- \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)

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

Simplify

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

Compute the boundaries:

- (- 1)^{n} \frac{π}{n}

= - (- (- 1)^{n} \frac{π}{n})

Simplify

= \frac{π ( - 1 ) ^{n}}{n}

\int_{- π}^{0} (t - π) sin (n t) d t

\int_{2}^{9} \frac{1}{8 x + 9} d x

\int_{- 1}^{0} e^{- 2 x} d x

\int_{0}^{r} (b \cdot r^{3} + c \cdot r^{2}) d r

\int_{0}^{1} tan (π x) d x

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