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Question Number 154669 by SANOGO last updated on 20/Sep/21

	Answered by ARUNG_Brandon_MBU last updated on 20/Sep/21

	$I = \int_{0}^{1} \frac{d t}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}} = \frac{1}{2} \int_{0}^{1} (\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}) d t$ $= \frac{1}{2} \int_{0}^{argsh (1)} \cosh^{2} ϑ d ϑ - \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos^{2} ϑ d ϑ$ $= \frac{1}{4} \int_{0}^{argsh (1)} (1 + \cosh 2 ϑ) d ϑ - \frac{1}{4} \int_{0}^{\frac{π}{2}} (\cos 2 ϑ + 1) d ϑ$ $= \frac{1}{4} argsh (1) + {[\frac{\sinh 2 ϑ}{8}]}_{0}^{argsh (1)} - \frac{1}{4} {[\frac{\sin 2 ϑ}{2} + ϑ]}_{0}^{\frac{π}{2}}$ $= \frac{1}{4} \ln (1 + \sqrt{2}) + \frac{\sqrt{2}}{4} - \frac{π}{8}$
	Commented by ARUNG_Brandon_MBU last updated on 20/Sep/21

	$Ah oui! merci .$
	Commented by puissant last updated on 20/Sep/21

	$B r o o a t t e n t i o n!!!!$ $\frac{1}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}} = \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{2 t^{2}} . .$
	Commented by SANOGO last updated on 20/Sep/21

	$s u p e r e d e m o n s t r a t i n$
	Answered by ARUNG_Brandon_MBU last updated on 20/Sep/21

	$I = \int_{0}^{1} \frac{d t}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}} = \int_{0}^{1} \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{2 t^{2}} d t$ $= \frac{1}{2} \int_{0}^{argsh (1)} \frac{\cosh^{2} ϑ}{\sinh^{2} ϑ} d ϑ - \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\cos^{2} ϑ}{\sin^{2} ϑ} d ϑ$ $= \frac{1}{2} \int_{0}^{argsh (1)} ({cosech}^{2} ϑ + 1) d ϑ - \frac{1}{2} \int_{0}^{\frac{π}{2}} (\csc^{2} ϑ - 1) d ϑ$ $= \frac{1}{2} {[ϑ - \coth ϑ]}_{0}^{argsh (1)} + \frac{1}{2} {[\cot ϑ + ϑ]}_{0}^{\frac{π}{2}}$ $= \frac{1}{2} \ln (1 + \sqrt{2}) - \frac{\sqrt{2}}{2} + \frac{π}{4}$
	Commented by SANOGO last updated on 20/Sep/21

	$m e r c i b i e n m o n g r a n d$
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