integral from 0 to pi of cos(x)*sin(nx)

解答

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

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

求解步骤

\int_{0}^{π} cos (x) sin (n x) d x

利用以下特性:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

\int_{0}^{π} sin (n x + x) d x = \frac{( - 1 ) ^{n} + 1}{n + 1}

\int_{0}^{π} sin (n x - x) d x = \frac{( - 1 ) ^{n} + 1}{n - 1}

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

化简

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

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

\int_{0}^{1} 2 t^{2} d t

\int_{1}^{9} \frac{e ^{\frac{9}{x}}}{x ^{2}} d x

\int_{2}^{5} - \frac{3}{x ^{2}} d x

\int_{1}^{3} \frac{2 x}{x ^{2} + 1} d x

\int_{0}^{1} (x + 1)^{2} d x