integral of (1+cos(2θ))/2

解答

\int \frac{1 + cos ( 2 θ )}{2} d θ

\frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) + C

求解步骤

\int \frac{1 + cos ( 2 θ )}{2} d θ

提出常数:

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

= \frac{1}{2} \cdot \int 1 + cos (2 θ) d θ

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 1 d θ + \int cos (2 θ) d θ)

\int 1 d θ = θ

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

= \frac{1}{2} (θ + \frac{1}{2} sin (2 θ))

解答补常数

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

\frac{\partial}{\partial s} (4 s t)

\frac{d y}{d x} = y - 2 x

x \to 4 lim (x - 6)

x \cdot \frac{d y}{d x} + 3 y = x^{3} - x

d er i v a t i v e f (x) = (6 x^{2} + 4 x)^{2}