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Category: Integration

sin2x-cos-4-x-1-dx-

Question Number 177775 by depressiveshrek last updated on 08/Oct/22 $$\int\frac{\mathrm{sin2}{x}}{\:\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}}{dx} \\ $$ Answered by Ar Brandon last updated on 08/Oct/22 $${I}=\int\frac{\mathrm{sin2}{x}}{\:\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}}{dx}=\int\frac{\mathrm{2sin}{x}\mathrm{cos}{x}}{\:\sqrt{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{1}}}{dx}\:,\:{c}=\mathrm{cos}{x} \\…

Question-46641

Question Number 46641 by Tinkutara last updated on 29/Oct/18 Commented by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18 $${question}\:{itself}\:\:{is}\:{doubtful}…{because} \\ $$$$\int\frac{{xdx}}{\:\sqrt{\mathrm{1}+\left({x}−\frac{\mathrm{1}}{{x}}\right)}}{dx}\:\leftarrow{it}\:{has}\:{no}\:{upper}\:{or}\:{lower}\:{limit} \\ $$ Terms of Service Privacy…

Question-46639

Question Number 46639 by Tinkutara last updated on 29/Oct/18 Commented by maxmathsup by imad last updated on 30/Oct/18 $${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}+\mathrm{1}}{dx}\:\:\:\Rightarrow{A}=_{\mathrm{2}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sin}\left({t}\right)}{\frac{{t}}{\mathrm{2}}+\mathrm{1}}\:\frac{{dt}}{\mathrm{2}} \\…