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受欢迎的微积分 >

integral from 0 to pi/4 of cos^5(2x)

解答

\int_{0}^{\frac{π}{4}} cos^{5} (2 x) d x

解答

\frac{4}{15}

+1

十进制

0.26666 \dots

求解步骤

\int_{0}^{\frac{π}{4}} cos^{5} (2 x) d x

使用换元积分法

= \int_{0}^{\frac{π}{2}} cos^{5} (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}} cos^{5} (u) d u

使用三角恒等式改写

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

使用换元积分法

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

乘开

(1 - v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= \frac{1}{2} \cdot \int_{0}^{1} 1 - 2 v^{2} + v^{4} 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}{2} (\int_{0}^{1} 1 d v - \int_{0}^{1} 2 v^{2} d v + \int_{0}^{1} v^{4} d v)

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

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

\int_{0}^{1} v^{4} d v = \frac{1}{5}

= \frac{1}{2} (1 - \frac{2}{3} + \frac{1}{5})

化简

\frac{1}{2} (1 - \frac{2}{3} + \frac{1}{5}) : \frac{4}{15}

= \frac{4}{15}

作图

流行的例子

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\int_{0}^{2} - 2 x + 4 d x

integral from 1 to 5 of 5x

\int_{1}^{5} 5 x d x

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\int_{0}^{14} 196 - x^{2} d x

integral from 0 to pi/4 of (1+e^{tan(x)})sec^2(x)

\int_{0}^{\frac{π}{4}} (1 + e^{t a n (x)}) sec^{2} (x) d x

integral from 1 to 6 of 2x-6

\int_{1}^{6} 2 x - 6 d x