integral of sin(4x)cos(5x)

解答

\int sin (4 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{9} cos (9 x) + cos (x)) + C

求解步骤

\int sin (4 x) cos (5 x) d x

利用以下特性:

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

= \int \frac{sin ( 4 x + 5 x ) + sin ( 4 x - 5 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 sin (4 x + 5 x) + sin (4 x - 5 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 sin (4 x + 5 x) d x + \int sin (4 x - 5 x) d x)

\int sin (4 x + 5 x) d x = - \frac{1}{9} cos (9 x)

\int sin (4 x - 5 x) d x = cos (x)

= \frac{1}{2} (- \frac{1}{9} cos (9 x) + cos (x))

解答补常数

= \frac{1}{2} (- \frac{1}{9} cos (9 x) + cos (x)) + C

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