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Category: Integration

x-1-2x-x-2-dx-

Question Number 59807 by aliesam last updated on 15/May/19 $$\int\frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }}\:{dx} \\ $$ Answered by tanmay last updated on 15/May/19 $${t}^{\mathrm{2}} =\mathrm{2}{x}−{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{tdt}=\left(\mathrm{2}−\mathrm{2}{x}\right){dx} \\…

Question-59803

Question Number 59803 by bhanukumarb2@gmail.com last updated on 15/May/19 Commented by maxmathsup by imad last updated on 16/May/19 $${let}\:{A}\:=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left({lnx}\right)^{\mathrm{2}} }{\:\sqrt{{x}}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:{dx}\:\Rightarrow\pi^{\mathrm{2}} \:{A}\:=_{\sqrt{{x}}={t}}…

find-the-general-solution-y-t-of-the-ordinary-differential-equation-y-2-y-cos-t-where-w-gt-0-

Question Number 59800 by necx1 last updated on 14/May/19 $${find}\:{the}\:{general}\:{solution}\:{y}\left({t}\right)\:{of}\:{the} \\ $$$${ordinary}\:{differential}\:{equation} \\ $$$${y}''\:+\:\omega^{\mathrm{2}} {y}=\mathrm{cos}\:\omega{t}\:,{where}\:{w}>\mathrm{0} \\ $$ Answered by MJS last updated on 15/May/19 $$\mathrm{1}^{\mathrm{st}}…

x-x-3-1-x-2-dx-

Question Number 125313 by john_santu last updated on 10/Dec/20 $$\:\:\:\beta\left({x}\right)=\int\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\: \\ $$ Answered by john_santu last updated on 10/Dec/20 $${let}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:=\:{w}\:\Rightarrow{x}^{\mathrm{2}} \:=\:\mathrm{1}−{w}^{\mathrm{2}} \\…

Question-190841

Question Number 190841 by Rupesh123 last updated on 12/Apr/23 Answered by ARUNG_Brandon_MBU last updated on 12/Apr/23 $${I}=\int_{\mathrm{0}} ^{\pi} \frac{{xdx}}{\mathrm{1}+\mathrm{cos}\alpha\mathrm{sin}{x}}=\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{\mathrm{1}+\mathrm{cos}\alpha\mathrm{sin}{x}}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{1}+\mathrm{cos}\alpha\mathrm{sin}{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}}…

Question-190812

Question Number 190812 by Mahliyo last updated on 12/Apr/23 Commented by Frix last updated on 12/Apr/23 $$\mathrm{Use}\:\mathrm{3}\:\mathrm{steps} \\ $$$$\mathrm{1}.\:{t}={x}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}.\:{u}=\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{2}{t}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{3}.\:{v}=\mathrm{tan}\:\frac{{u}}{\mathrm{2}} \\…