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受欢迎的微积分 >

integral from 0 to 1 of pi(tan(pi/4 y))^2

解答

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

解答

- π + 4

+1

十进制

0.85840 \dots

求解步骤

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

提出常数:

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

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

化简

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

使用换元积分法

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

提出常数:

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

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

使用三角恒等式改写

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

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

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

\int_{0}^{\frac{π}{4}} 1 d u = \frac{π}{4}

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

= π \frac{4}{π} (- \frac{π}{4} + 1)

化简

π \frac{4}{π} (- \frac{π}{4} + 1) : - π + 4

= - π + 4

作图

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