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

The integral from-pi to pi of xsin(x)cos(nx) is (2pi(-1)^{-n})/((n+1)(-n+1))

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integral from-pi to pi of xsin(x)cos(nx)

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

\frac{2 π ( - 1 ) ^{- n}}{( n + 1 ) ( - n + 1 )}

Solution steps

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

Apply Integration By Parts

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

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

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

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

2 \cdot \frac{1}{2} (- \frac{1}{( 1 + n ) ^{2}} sin (x + n x) - \frac{1}{( 1 - n ) ^{2}} sin (x - n x)) (n + 1) (- n + 1) = (- \frac{1}{( 1 + n ) ^{2}} sin (x + n x) - \frac{1}{( 1 - n ) ^{2}} sin (x - n x)) (n + 1) (- n + 1)

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

Compute the boundaries:

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

= 2 \cdot \frac{π ( - 1 ) ^{- n}}{( n + 1 ) ( - n + 1 )}

Simplify

= \frac{2 π ( - 1 ) ^{- n}}{( n + 1 ) ( - n + 1 )}

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

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

\int_{2}^{6} 2 d x

\int_{- 2 π}^{2 π} 8 sin (x) d x

\int_{π}^{2 π} sin^{2} (θ) d θ

What is the integral from-pi to pi of xsin(x)cos(nx) ?
The integral from-pi to pi of xsin(x)cos(nx) is (2pi(-1)^{-n})/((n+1)(-n+1))