integral of sin(x)sin(x+pi/6)

解答

\int sin (x) sin (x + \frac{π}{6}) d x

\frac{1}{4} (- sin (2 x + \frac{π}{6}) + 3 x) + C

求解步骤

\int sin (x) sin (x + \frac{π}{6}) d x

使用三角恒等式改写

= \int \frac{1}{4} (- 2 cos (2 x + \frac{π}{6}) + 3) d x

提出常数:

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

= \frac{1}{4} \cdot \int - 2 cos (2 x + \frac{π}{6}) + 3 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}{4} (- \int 2 cos (2 x + \frac{π}{6}) d x + \int 3 d x)

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

\int 3 d x = 3 x

= \frac{1}{4} (- sin (2 x + \frac{π}{6}) + 3 x)

解答补常数

= \frac{1}{4} (- sin (2 x + \frac{π}{6}) + 3 x) + C

\frac{\partial}{\partial x} (6 x^{2} cos (y^{7}))

\int_{1}^{5} 5 d x

x \to 0 + lim (x^{3})

in v erse l a pl a ce \frac{s}{( s + 2 ) ( s ^{2} + 4 )}

\int (x + 3 x) d x