Question Number 82566 by jagoll last updated on 22/Feb/20 $$\int\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:{dx}\:=\:? \\ $$ Commented by mathmax by abdo last updated on 22/Feb/20 $${let}\:{I}=\int\sqrt{\frac{{x}+\mathrm{1}}{{x}}}{dx}\:{changement}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}={t}\:{give}\:\frac{{x}+\mathrm{1}}{{x}}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}+\mathrm{1}\:={xt}^{\mathrm{2}}…
Question Number 82560 by niroj last updated on 22/Feb/20 $$\:\boldsymbol{\mathrm{U}}\mathrm{se}\:\mathrm{gamma}\:\mathrm{function}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\:\:\left(\mathrm{i}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}\:\mathrm{2}\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{dx}}\:=\:\frac{\mathrm{3}\boldsymbol{\pi}−\mathrm{4}}{\mathrm{192}}. \\ $$$$\:\:\left(\mathrm{ii}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\boldsymbol{\mathrm{cos}}^{\mathrm{4}} \mathrm{3}\boldsymbol{\theta}\:\mathrm{sin}^{\mathrm{2}} \mathrm{6}\theta\:=\:\frac{\mathrm{5}\pi}{\mathrm{192}}. \\ $$ Answered by…
Question Number 17010 by arnabpapu550@gmail.com last updated on 29/Jun/17 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\right)^{\mathrm{2}} \mathrm{dx} \\ $$ Answered by sma3l2996 last updated on 29/Jun/17 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 17011 by arnabpapu550@gmail.com last updated on 29/Jun/17 $$\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{x}\sqrt{\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$ Answered by sma3l2996 last updated on 29/Jun/17…
Question Number 17008 by arnabpapu550@gmail.com last updated on 29/Jun/17 $$\int_{\frac{\mathrm{1}\:}{\Pi}} ^{\frac{\mathrm{2}}{\Pi}} \:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{sin}\frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx} \\ $$ Answered by sma3l2996 last updated on 29/Jun/17 $${t}=\frac{\mathrm{1}}{{x}}\Rightarrow{dt}=\frac{−{dx}}{{x}^{\mathrm{2}} } \\…
Question Number 17009 by arnabpapu550@gmail.com last updated on 29/Jun/17 $$\int_{\mathrm{0}} ^{\:\mathrm{a}} \:\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$ Commented by prakash jain last updated on 29/Jun/17…
Question Number 82531 by Power last updated on 22/Feb/20 Commented by abdomathmax last updated on 24/Feb/20 $${I}\:=\int\:\frac{\sqrt{\mathrm{1}−{x}}}{{x}}{dx}\:+\int\:\frac{\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{{x}}{dx}\:+\int\:\:\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{4}} }}{{x}}{dx}\:{we}\:{have} \\ $$$$\int\:\:\frac{\sqrt{\mathrm{1}−{x}}}{{x}}{dx}\:\:{changement}\:\sqrt{\mathrm{1}−{x}}={t}\:{give}\mathrm{1}−{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\mathrm{1}−{t}^{\mathrm{2}} \:\Rightarrow\int\:\:\frac{\sqrt{\mathrm{1}−{x}}}{{x}}{dx}\:=\:\int\:\:\frac{{t}}{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 82510 by jagoll last updated on 22/Feb/20 $$\int\:\sqrt{{x}+\sqrt{{x}}\:}\:{dx}\:=\:? \\ $$ Commented by mathmax by abdo last updated on 23/Feb/20 $${I}\:=\int\sqrt{{x}+\sqrt{{x}}}{dx}\:{changement}\:\sqrt{{x}}={t}\:{give}\:{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\sqrt{{t}^{\mathrm{2}}…
Question Number 148026 by mnjuly1970 last updated on 25/Jul/21 $$ \\ $$$$\mathrm{A}\::=\int_{−\mathrm{1}} ^{\:\mathrm{0}} {e}^{\:{x}\:+\frac{\mathrm{1}}{{x}}} \:\:\&\:{a}\mathrm{A}+{e}^{{b}} =\:\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{{e}}\right)^{\:{x}+\frac{\mathrm{1}}{{x}}} \\ $$$$\:\:{than}\::\:\:{a}+\:{b}\:=?\:\:\:\:\:{a}\:,\:{b}\:\in\:\mathbb{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered…
Question Number 148020 by qaz last updated on 25/Jul/21 $$\underset{\mathrm{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \centerdot\frac{\mathrm{lnk}}{\mathrm{k}}=\gamma\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{2} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com