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受欢迎的微积分 >

integral from 0 to pi/4 of sin^3(4x)

解答

\int_{0}^{\frac{π}{4}} sin^{3} (4 x) d x

解答

\frac{1}{3}

+1

十进制

0.33333 \dots

求解步骤

\int_{0}^{\frac{π}{4}} sin^{3} (4 x) d x

使用换元积分法

= \int_{0}^{π} sin^{3} (u) \frac{1}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int_{0}^{π} sin^{3} (u) d u

使用三角恒等式改写

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

使用换元积分法

= \frac{1}{4} \cdot \int_{1}^{- 1} - 1 + v^{2} d v

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

= \frac{1}{4} (- \int_{- 1}^{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}{4} (- (- \int_{- 1}^{1} 1 d v + \int_{- 1}^{1} v^{2} d v))

\int_{- 1}^{1} 1 d v = 2

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

= \frac{1}{4} (- (- 2 + \frac{2}{3}))

化简

\frac{1}{4} (- (- 2 + \frac{2}{3})) : \frac{1}{3}

= \frac{1}{3}

作图

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