每日一题[2465]积分

$\displaystyle\int_0^{\frac{\pi}2}\dfrac1{1+\cos^2t}{ {\rm d}} t=$_______.

答案    $\dfrac{\sqrt2\pi}{4}$.

解析    根据题意, 有\[\begin{split} \int_0^{\frac{\pi}2}\dfrac1{1+\cos^2t}{ {\rm d}} t&=\int_0^{\frac{\pi}2}\dfrac1{\sin^2t+2\cos^2t}{ {\rm d}} t\\ &=\int_0^{\frac{\pi}2}\dfrac{1}{\tan^2t+2}{ {\rm d}}(\tan t)\\ &=\dfrac{1}{\sqrt 2}\int_0^{\frac{\pi}2}\dfrac{1}{\left(\dfrac{\tan t}{\sqrt 2}\right)^2+1}{ {\rm d}}\left(\dfrac{\tan t}{\sqrt 2}\right)\\ &=\dfrac{1}{\sqrt 2}\arctan\dfrac{\tan t}{\sqrt 2}\Big|_{0}^{\frac{\pi}2}\\ &=\dfrac{\sqrt2\pi}{4}.\end{split}\]

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