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Category: Integration

dx-tan-4-x-1-

Question Number 125895 by john_santu last updated on 15/Dec/20 $$\:\:\:\:\int\:\frac{{dx}}{\mathrm{tan}\:^{\mathrm{4}} {x}+\mathrm{1}}\:? \\ $$ Answered by Dwaipayan Shikari last updated on 15/Dec/20 $$\int\frac{{dx}}{{tan}^{\mathrm{4}} {x}+\mathrm{1}}\:\:\:\:\:\:\:\:\:{tanx}={t}\Rightarrow{sec}^{\mathrm{2}} {x}=\frac{{dt}}{{dx}} \\…

tan-3-x-sec-x-dx-

Question Number 125889 by bramlexs22 last updated on 15/Dec/20 $$\:\:\int\:\frac{\mathrm{tan}\:^{\mathrm{3}} {x}}{\:\sqrt{\mathrm{sec}\:{x}}}\:{dx}\:?\: \\ $$ Answered by bobhans last updated on 15/Dec/20 $$\int\:\frac{\mathrm{tan}\:{x}\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}\right)}{\:\sqrt{\mathrm{sec}\:{x}}}\:{dx}\:= \\ $$$$\:\left[\:\mathrm{sec}\:{x}\:=\:{u}^{\mathrm{2}} \:\Rightarrow\mathrm{tan}\:{x}\:{dx}\:=\:\frac{\mathrm{2}\:{du}}{{u}}\:\right]…

Question-125876

Question Number 125876 by bramlexs22 last updated on 14/Dec/20 Commented by liberty last updated on 15/Dec/20 $${W}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{Lim}}\:\underset{{i}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\rho{A}\left({x}_{{i}} ^{\ast} \right)\Delta{x}\:\:\:\:\:\:\: \\ $$$${W}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{Lim}}\:\underset{{i}\:=\:\mathrm{1}}…

Question-60320

Question Number 60320 by Sardor2211 last updated on 19/May/19 Commented by maxmathsup by imad last updated on 20/May/19 $${let}\:{I}\:=\int\:\:\frac{{x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} \:+\mathrm{5}{x}\:+\mathrm{23}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−\mathrm{5}\right)}{dx}\:\:{let}\:{decompose}\:{F}\left({x}\right)\:=\frac{{x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} \:+\mathrm{5}{x}\:+\mathrm{23}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−\mathrm{5}\right)}…

Question-60318

Question Number 60318 by Sardor2211 last updated on 19/May/19 Commented by kaivan.ahmadi last updated on 20/May/19 $${we}\:{find}\:{I}=\int{x}^{\mathrm{2}} {arctgx}\:{dx}\:{then}\:{multiply}\:\mathrm{4} \\ $$$${u}={arctgx}\Rightarrow{du}=\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${dv}={x}^{\mathrm{2}} {dx}\Rightarrow{v}=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}…