integral of-4cos^2(x)

解答

\int - 4 cos^{2} (x) d x

- 2 x - sin (2 x) + C

求解步骤

\int - 4 cos^{2} (x) d x

提出常数:

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

= - 4 \cdot \int cos^{2} (x) d x

使用三角恒等式改写

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

提出常数:

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

= - 4 \cdot \frac{1}{2} \cdot \int 1 + 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

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

\int 1 d x = x

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

= - 4 \cdot \frac{1}{2} (x + \frac{1}{2} sin (2 x))

化简

- 4 \cdot \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : - 2 x - sin (2 x)

= - 2 x - sin (2 x)

解答补常数

= - 2 x - sin (2 x) + C

\frac{d}{d x} ((x - 1) (x + 2))

d er i v a t i v e \frac{1}{7} π r^{3}

\frac{\partial}{\partial x} (ln (3 x))

x \to \infty lim (8 x^{2} e^{- x})

y^{^{''}} - y^{^{'}} + y = e^{2 t}