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Category: Integration

calculate-dx-2-x-2-3-x-2-

Question Number 138128 by mathmax by abdo last updated on 10/Apr/21 $$\mathrm{calculate}\:\int\:\:\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{2}−\mathrm{x}^{\mathrm{2}} }+\sqrt{\mathrm{3}+\mathrm{x}^{\mathrm{2}} }} \\ $$ Answered by MJS_new last updated on 10/Apr/21 $$\int\frac{{dx}}{\:\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{3}+{x}^{\mathrm{2}}…

calculus-III-evaluate-0-pi-2-0-x-cos-y-pi-2-x-pi-2-y-dydx-

Question Number 138114 by mnjuly1970 last updated on 10/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:……\:{calculus}…..\left({III}\right)…… \\ $$$$\:\:\:\:\:\:\:\:\:{evaluate}::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\overset{???} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:{x}} \frac{{cos}\left({y}\right)}{\:\sqrt{\left(\frac{\pi}{\mathrm{2}}−{x}\right)\left(\frac{\pi}{\mathrm{2}}−{y}\right)}}{dydx} \\ $$$$ \\ $$ Commented by…

Question-7026

Question Number 7026 by Tawakalitu. last updated on 07/Aug/16 Commented by Yozzii last updated on 07/Aug/16 $${Write}\:\int\frac{{sinx}+{cosx}}{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}}{dx}=\int\frac{{sinx}}{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}}{dx}+\int\frac{{cosx}}{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}}{dx}. \\ $$$${Evaluate}\:{each}\:{integral}\:{by}\:{the}\:{following}\:{steps}…

0-pi-2-ln-ln-2-sin-pi-2-ln-2-sin-ln-cos-tan-d-

Question Number 138086 by EnterUsername last updated on 10/Apr/21 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{sin}\theta\right)}{\pi^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{sin}\theta\right)}\right)\:\frac{\mathrm{ln}\left(\mathrm{cos}\theta\right)}{\mathrm{tan}\theta}\mathrm{d}\theta \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

nice-calculus-prove-n-1-2n-1-1-n-1-log-2-

Question Number 138065 by mnjuly1970 last updated on 09/Apr/21 $$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:…{nice}\:…\:…\:…\:{calculus}… \\ $$$$\:\:\:\:\:\:\:{prove}:: \\ $$$$\:\:\:\:\:\:\:\:\:\Omega=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{1}}{{n}+\mathrm{1}}\:=−\gamma+{log}\left(\mathrm{2}\right) \\ $$ Answered by Ñï= last updated…