zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral from 0 to pi/2 of sin^3(3x)

解答

\int_{0}^{\frac{π}{2}} sin^{3} (3 x) d x

解答

\frac{2}{9}

+1

十进制

0.22222 \dots

求解步骤

\int_{0}^{\frac{π}{2}} sin^{3} (3 x) d x

使用换元积分法

= \int_{0}^{\frac{3 π}{2}} sin^{3} (u) \frac{1}{3} d u

提出常数:

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

= \frac{1}{3} \cdot \int_{0}^{\frac{3 π}{2}} sin^{3} (u) d u

使用三角恒等式改写

= \frac{1}{3} \cdot \int_{0}^{\frac{3 π}{2}} (1 - cos^{2} (u)) sin (u) d u

使用换元积分法

= \frac{1}{3} \cdot \int_{1}^{0} - 1 + v^{2} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{3} (- \int_{0}^{1} - 1 + v^{2} d v)

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}{3} (- (- \int_{0}^{1} 1 d v + \int_{0}^{1} v^{2} d v))

\int_{0}^{1} 1 d v = 1

\int_{0}^{1} v^{2} d v = \frac{1}{3}

= \frac{1}{3} (- (- 1 + \frac{1}{3}))

化简

\frac{1}{3} (- (- 1 + \frac{1}{3})) : \frac{2}{9}

= \frac{2}{9}

作图

流行的例子

tangent (2x-1)^2ln(2x-1)

t an g e n t (2 x - 1)^{2} ln (2 x - 1)

(dy)/(dx)=y(1-y),y(0)=2

\frac{d y}{d x} = y (1 - y), y (0) = 2

derivative of (-2x^2+2/((1+x^2)^2))

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

(\partial)/(\partial x)(-4y(-2x-y-3))

\frac{\partial}{\partial x} (- 4 y (- 2 x - y - 3))

integral of 1-(16)/(sin^4(x))

\int 1 - \frac{16}{sin ^{4} ( x )} d x