integral of cos(4x)cos(2x)

解答

\int cos (4 x) cos (2 x) d x

\frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x)) + C

求解步骤

\int cos (4 x) cos (2 x) d x

使用三角恒等式改写

= \int \frac{1}{2} (cos (6 x) + cos (2 x)) 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 (6 x) + cos (2 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 cos (6 x) d x + \int cos (2 x) d x)

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

= \frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x))

解答补常数

= \frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x)) + C

\frac{d}{d x} (a r (x) cos (x))

\int \frac{- x - 1}{x ^{2} + 3} d x

\int x x^{\frac{3}{2}} - 1 d x

a re a sin (x), cos (x), x = - \frac{π}{6}

\frac{d}{d x} (x^{2} e^{5 x})