integral from pi/3 to pi/6 of tan^2(x)

解答

\int_{\frac{π}{3}}^{\frac{π}{6}} tan^{2} (x) d x

\frac{π}{6} + \frac{1}{3} - 3

十进制

- 0.63110 \dots

求解步骤

\int_{\frac{π}{3}}^{\frac{π}{6}} tan^{2} (x) d x

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

= - \int_{\frac{π}{6}}^{\frac{π}{3}} tan^{2} (x) d x

使用三角恒等式改写

= - \int_{\frac{π}{6}}^{\frac{π}{3}} - 1 + sec^{2} (x) d x

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

= - (- \int_{\frac{π}{6}}^{\frac{π}{3}} 1 d x + \int_{\frac{π}{6}}^{\frac{π}{3}} sec^{2} (x) d x)

\int_{\frac{π}{6}}^{\frac{π}{3}} 1 d x = \frac{π}{6}

\int_{\frac{π}{6}}^{\frac{π}{3}} sec^{2} (x) d x = 3 - \frac{1}{3}

= - (- \frac{π}{6} + 3 - \frac{1}{3})

化简

= \frac{π}{6} + \frac{1}{3} - 3

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