integral of 9cos^2(θ)

解答

\int 9 cos^{2} (θ) d θ

\frac{9}{2} (θ + \frac{1}{2} sin (2 θ)) + C

求解步骤

\int 9 cos^{2} (θ) d θ

提出常数:

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

= 9 \cdot \int cos^{2} (θ) d θ

使用三角恒等式改写

= 9 \cdot \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

= 9 \cdot \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

= 9 \cdot \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)

= 9 \cdot \frac{1}{2} (θ + \frac{1}{2} sin (2 θ))

化简

9 \cdot \frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) : \frac{9}{2} (θ + \frac{1}{2} sin (2 θ))

= \frac{9}{2} (θ + \frac{1}{2} sin (2 θ))

解答补常数

= \frac{9}{2} (θ + \frac{1}{2} sin (2 θ)) + C

d er i v a t i v e [3 (3 x^{5} + 6)^{3}] (12 x^{3})

a re a x = 9 y^{3}, x = 23 y^{2}

\frac{\partial}{\partial z} (e^{x y})

d er i v a t i v e f (x) = (2 x + 1)^{5}

d er i v a t i v e \frac{( 1 - x )}{( x + 2 )}