Menu Close

Author: Tinku Tara

Question-82330

Question Number 82330 by Power last updated on 20/Feb/20 Commented by mathmax by abdo last updated on 20/Feb/20 $${let}\:{I}\:=\int{x}^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx}\:\:{changement}\:{x}={ash}\left({t}\right)\:{give} \\ $$$${I}\:=\int\:{a}^{\mathrm{2}} {sh}^{\mathrm{2}}…

Question-16794

Question Number 16794 by tawa tawa last updated on 26/Jun/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17 $${c}−{x}={t}^{\mathrm{3}} ,{x}−{a}={s}^{\mathrm{3}} \Rightarrow{x}=\frac{{s}^{\mathrm{3}} −{t}^{\mathrm{3}} }{\mathrm{2}}+\frac{{a}+{c}}{\mathrm{2}} \\ $$$${a}+{c}={c}−\mathrm{2}+{c}=\mathrm{2}\left({c}−\mathrm{1}\right)=\mathrm{2}{b},{c}−{a}=\mathrm{2} \\…

secx-dx-

Question Number 147867 by Khalmohmmad last updated on 24/Jul/21 $$\int\mathrm{sec}{x}\:{dx}=? \\ $$ Answered by Olaf_Thorendsen last updated on 24/Jul/21 $$\mathrm{F}\left({x}\right)\:=\:\int\mathrm{sec}{x}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{{dx}}{\mathrm{cos}{x}} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}} \\…

Question-16789

Question Number 16789 by ajfour last updated on 26/Jun/17 Commented by ajfour last updated on 26/Jun/17 $$\:\mathrm{solution}\:\mathrm{to}\:\mathrm{Q}.\mathrm{16065} \\ $$$$\mathrm{find}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{M}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{Area}\left(\bigtriangleup\mathrm{MAB}\right)=\mathrm{2Area}\left(\bigtriangleup\mathrm{MCD}\right). \\ $$ Answered by…

Question-16788

Question Number 16788 by jyoti5726 last updated on 26/Jun/17 Commented by 1234Hello last updated on 04/Jul/17 $$\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{3}}\:=\:\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{sec}^{−\mathrm{1}} \:\left(−\mathrm{2}\right)\:=\:\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{3}}\:−\:\mathrm{sec}^{−\mathrm{1}}…

Question-16785

Question Number 16785 by RasheedSoomro last updated on 26/Jun/17 Commented by RasheedSoomro last updated on 26/Jun/17 $$\mathrm{Without}\:\mathrm{using}\:\mathrm{area}-\mathrm{formula}\:\mathrm{find}\:\mathrm{out} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part}\:\mathrm{to}\:\mathrm{the}\:\mathrm{whole}. \\ $$ Commented by mrW1 last…