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受欢迎的微积分 >

integral from 0 to pi/4 of sin^4(2x)

解答

\int_{0}^{\frac{π}{4}} sin^{4} (2 x) d x

解答

\frac{3 π}{32}

+1

十进制

0.29452 \dots

求解步骤

\int_{0}^{\frac{π}{4}} sin^{4} (2 x) d x

使用换元积分法

= \int_{0}^{\frac{π}{2}} sin^{4} (u) \frac{1}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) d u

使用降次积分法

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u)

\int_{0}^{\frac{π}{2}} sin^{2} (u) d u = \frac{π}{4}

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4})

化简

\frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4}) : \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

= \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

带入上下限计算后得:

0

= \frac{1}{2} (- 0 + \frac{3 π}{16})

化简

= \frac{3 π}{32}

作图

流行的例子

derivative 5/(x^4)

d er i v a t i v e \frac{5}{x ^{4}}

derivative (2x)/(5-cot(x))

d er i v a t i v e \frac{2 x}{5 - cot ( x )}

derivative of ln^2(5x)

\frac{d}{d x} (ln^{2} (5 x))

limit as (x,y) approaches (2,3) of x+y^2

(x, y) \to (2, 3) lim (x + y^{2})

tangent f(x)=2x^3+42x^2+270x+4

t an g e n t f (x) = 2 x^{3} + 42 x^{2} + 270 x + 4