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Author: Tinku Tara

I-cosx-2sinx-e-2x-sinx-dx-

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}}…

log-e-1-cox-1-cosx-differentiate-w-r-t-x-

Question Number 19860 by vivek last updated on 16/Aug/17 $$\mathrm{log}_{{e}} \sqrt{\frac{\mathrm{1}−{cox}}{\mathrm{1}+{cosx}}\:\:\boldsymbol{{differentiate}}\:\boldsymbol{{w}}.\boldsymbol{{r}}.\boldsymbol{{t}}\:\boldsymbol{{x}}} \\ $$ Answered by myintkhaing last updated on 17/Aug/17 $$\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{log}_{\mathrm{e}} \:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right) \\ $$$$=\:\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\right)\left(\frac{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\left(−\mathrm{sin}\:\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }\right)…

1-x-1-x-1-x-1-x-differentiate-w-r-t-x-

Question Number 19859 by vivek last updated on 16/Aug/17 $$\frac{\sqrt{}\mathrm{1}+{x}−\sqrt{}\mathrm{1}−{x}}{\:\sqrt{}\mathrm{1}+{x}−\sqrt{}\mathrm{1}−{x}}\:\:\boldsymbol{{differentiate}}\:\boldsymbol{{w}}.\boldsymbol{{r}}.\boldsymbol{{t}}\:\boldsymbol{{x}} \\ $$ Commented by giridhar.mathmania last updated on 17/Aug/17 $$\mathrm{function}\:\mathrm{to}\:\mathrm{be}\:\mathrm{differentiated}\:\mathrm{is}\:\mathrm{1}. \\ $$$$\mathrm{hence}\:\mathrm{its}\:\mathrm{differential}\:\mathrm{is}\:\mathrm{0}. \\ $$ Terms…

Question-85395

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-85391

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…