问题113: 计算定积分∫tan2θ^2dθ

$$ \begin{aligned} & \quad \int_0^{\frac{\pi}{8}} \tan ^2 2 \theta d \theta \\ & = \frac{1}{2} \int_0^{\frac{\pi}{4}} \tan ^2 t d t (2\theta = t ) \\ & =\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{1-\cos ^2 t}{\cos ^2 t} d t \\ & =\frac{1}{2} \int_0^{\frac{\pi}{4}} \sec ^2 t d t-\frac{1}{2} \int_0^{\frac{\pi}{4}} 1 d t \\ & =\left.\frac{1}{2} \tan t\right|_0 ^{\frac{\pi}{4}}-\left.\frac{1}{2} t\right|_0 ^{\frac{\pi}{4}} \\ & =\frac{1}{2}\left(1-\frac{\pi}{4}\right) \\ & =\frac{4-\pi}{8} \\ & \end{aligned} $$

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最后修改于1月21日