Question Number 150976 by mnjuly1970 last updated on 17/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85441 by jagoll last updated on 22/Mar/20 $$\int\:\frac{\mathrm{6x}^{\mathrm{4}} −\mathrm{4}}{\:\sqrt{\mathrm{x}^{\mathrm{4}} −\mathrm{2}}}\:\mathrm{dx}\:=\:? \\ $$ Answered by john santu last updated on 22/Mar/20 $${I}\:=\:\int\:\frac{\mathrm{4}{x}^{\mathrm{4}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{2}}}\:{dx}\:=\:\mathrm{4}\int\:{x}\:\left(\frac{{x}^{\mathrm{3}}…
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Question Number 85426 by john santu last updated on 22/Mar/20 $$\int\:_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{9}+\mathrm{16sin}\:\mathrm{2}{x}}\:{dx} \\ $$ Commented by som(math1967) last updated on 22/Mar/20 $$\int\frac{{sinx}+{cosx}}{\mathrm{25}−\mathrm{16}\left(\mathrm{1}−{sin}\mathrm{2}{x}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{4}\left({sinx}+{cosx}\right)}{\mathrm{5}^{\mathrm{2}}…
Question Number 85414 by M±th+et£s last updated on 21/Mar/20 $$\int\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{cos}\left({x}\right)}\:}\:{dx} \\ $$ Commented by M±th+et£s last updated on 21/Mar/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\sqrt{{cos}\left({x}\right)}}\:{dx}\:\:\:{typo}….\: \\ $$ Answered…
Question Number 150932 by maged last updated on 16/Aug/21 $$\mathrm{I}=\int\frac{\mathrm{cosx}\:−\:\mathrm{2sinx}}{\mathrm{e}^{\mathrm{2x}} −\mathrm{sinx}}\mathrm{dx}\overset{?} {=} \\ $$ Answered by Olaf_Thorendsen last updated on 16/Aug/21 $$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{cos}{x}−\mathrm{2sin}{x}}{{e}^{\mathrm{2}{x}} −\mathrm{sin}{x}}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\left(\mathrm{2}{e}^{\mathrm{2}{x}}…
Question Number 150941 by rexford last updated on 16/Aug/21 $$\int_{\mathrm{1}} ^{\mathrm{4}} \mid{x}−\mathrm{2}\mid{dx} \\ $$$${please}\:{help}\:{me}\:{out} \\ $$ Commented by puissant last updated on 25/Aug/21 $$=\int_{\mathrm{1}} ^{\mathrm{4}}…
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Question Number 85395 by Power last updated on 21/Mar/20 Commented by MJS last updated on 21/Mar/20 $$\mathrm{let}\:{t}=\sqrt{\mathrm{ctg}\:{x}}\:\rightarrow\:{dx}=−\mathrm{2sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{ctg}\:{x}} \\ $$$$\Rightarrow\:\int\sqrt{\mathrm{ctg}\:{x}}{dx}=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\…
Question Number 85391 by sakeefhasan05@gmail.com last updated on 21/Mar/20 Commented by sakeefhasan05@gmail.com last updated on 21/Mar/20 $$\mathrm{I}\:\mathrm{have}\:\mathrm{got}\:\left(\frac{\mathrm{8}}{\mathrm{63}}\right)\:\mathrm{as}\:\mathrm{answer}\: \\ $$$$\mathrm{whoever}\:\mathrm{check}\:\mathrm{it}\:\mathrm{pls}\: \\ $$ Commented by mathmax by…