integral of cos(2x+9)cos(2x-9)

解答

\int cos (2 x + 9) cos (2 x - 9) d x

\frac{1}{2} (\frac{1}{4} sin (4 x) + cos (18) x) + C

求解步骤

\int cos (2 x + 9) cos (2 x - 9) d x

使用三角恒等式改写

= \int \frac{1}{2} (cos (4 x) + cos (18)) 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 cos (4 x) + cos (18) 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 cos (4 x) d x + \int cos (18) d x)

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

\int cos (18) d x = cos (18) x

= \frac{1}{2} (\frac{1}{4} sin (4 x) + cos (18) x)

解答补常数

= \frac{1}{2} (\frac{1}{4} sin (4 x) + cos (18) x) + C

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