Your Daily Experience of Math Adventures
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=$\dfrac{1}{\sqrt{2}}\int_{0}^{\dfrac{p}{4}}\dfrac{1}{sin(x+\dfrac{p}{4})}dx$=$\dfrac{1}{\sqrt{2}}\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}\dfrac{1}{sinu}du$=$\dfrac{1}{\sqrt{2}}\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}\dfrac{sin^{2}\dfrac{u}{2}+cos^{2}\dfrac{u}{2}}{2sin\dfrac{u}{2}cos\dfrac{u}{2}}du$=$\dfrac{1}{2\sqrt{2}}(\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}tan\dfrac{u}{2}du+\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}cot\dfrac{u}{2}du)$=$-\dfrac{1}{\sqrt{2}}ln(\sqrt{2}-1)$.
1 σχόλιο:
=$\dfrac{1}{\sqrt{2}}\int_{0}^{\dfrac{p}{4}}\dfrac{1}{sin(x+\dfrac{p}{4})}dx$=
ΑπάντησηΔιαγραφή$\dfrac{1}{\sqrt{2}}\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}\dfrac{1}{sinu}du$=
$\dfrac{1}{\sqrt{2}}\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}\dfrac{sin^{2}\dfrac{u}{2}+cos^{2}\dfrac{u}{2}}{2sin\dfrac{u}{2}cos\dfrac{u}{2}}du$=
$\dfrac{1}{2\sqrt{2}}(\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}tan\dfrac{u}{2}du+\int_{\dfrac{p}{4}}^{\dfrac{p}{2}}cot\dfrac{u}{2}du)$=
$-\dfrac{1}{\sqrt{2}}ln(\sqrt{2}-1)$.