integral from-pi to pi of sin(mx)sin(nx)

解答

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

- \frac{1}{m + n} sin (πm + πn) + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{2 ( m - n )}

求解步骤

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

使用三角恒等式改写

= \int_{- π}^{π} \frac{- cos ( m x + n x ) + cos ( m x - n x )}{2} d x

提出常数:

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

= \frac{1}{2} \cdot \int_{- π}^{π} - cos (m x + n x) + cos (m x - n x) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int_{- π}^{π} cos (m x + n x) d x + \int_{- π}^{π} cos (m x - n x) d x)

\int_{- π}^{π} cos (m x + n x) d x = \frac{2 sin ( πm + πn )}{m + n}

\int_{- π}^{π} cos (m x - n x) d x = \frac{sin ( πm - πn ) - sin ( - πm + πn )}{m - n}

= \frac{1}{2} (- \frac{2 sin ( πm + πn )}{m + n} + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{m - n})

化简

\frac{1}{2} (- \frac{2 sin ( πm + πn )}{m + n} + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{m - n}) : - \frac{1}{m + n} sin (πm + πn) + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{2 ( m - n )}

= - \frac{1}{m + n} sin (πm + πn) + \frac{sin ( πm - πn ) - sin ( - πm + πn )}{2 ( m - n )}

(y^{2} + 4) d x - \frac{1}{( 2 x + 5 ) ^{2}} d y = 0

\int_{0}^{1} \frac{x ^{2}}{25 + x ^{2}} d x

\int \frac{3}{t ^{4}} cos (\frac{1}{t ^{3}} - 9) d t

\int_{- 8}^{8} \frac{1}{2} (64 - x^{2}) d x

n = 1 \sum \infty \frac{8}{n ^{4}}