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Author: Tinku Tara

cothx-dx-

Question Number 76469 by kaivan.ahmadi last updated on 27/Dec/19 $$\int{cothx}\:{dx} \\ $$ Commented by mathmax by abdo last updated on 27/Dec/19 $$\int\:{coth}\left({x}\right){dx}\:=\int\:\frac{{ch}\left({x}\right)}{{sh}\left({x}\right)}{dx}\:=\int\:\:\frac{{e}^{{x}} \:+{e}^{−{x}} }{{e}^{{x}} −{e}^{−{x}}…

dx-e-2x-4-sec-1-e-x-4-

Question Number 76467 by kaivan.ahmadi last updated on 27/Dec/19 $$\int\frac{{dx}}{\:\sqrt{{e}^{\mathrm{2}{x}} −\mathrm{4}}\left({sec}^{−\mathrm{1}} \left(\frac{{e}^{{x}} }{\mathrm{4}}\right)\right)} \\ $$ Answered by john santu last updated on 28/Dec/19 $${let}\::\mathrm{sec}\:{u}\:=\:\frac{{e}^{{x}} }{\mathrm{4}}\:\rightarrow\:{e}^{{x}}…

cos3x-e-2x-dx-

Question Number 76462 by kaivan.ahmadi last updated on 27/Dec/19 $$\int{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} {dx} \\ $$ Commented by mathmax by abdo last updated on 27/Dec/19 $${let}\:{A}\:=\int\:{cos}\left(\mathrm{3}{x}\right){e}^{−\mathrm{2}{x}} {dx}\:\Rightarrow{A}\:={Re}\left(\int\:{e}^{\mathrm{3}{ix}} \:{e}^{−\mathrm{2}{x}}…

0-pi-2-sin-40x-sin-5x-dx-

Question Number 141998 by iloveisrael last updated on 25/May/21 $$\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{sin}\:\left(\mathrm{40}{x}\right)}{\mathrm{sin}\:\left(\mathrm{5}{x}\right)}\:{dx}\: \\ $$ Answered by EDWIN88 last updated on 25/May/21 $$\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{\mathrm{sin}\:\left(\mathrm{40x}\right)}{\mathrm{sin}\:\left(\mathrm{5x}\right)}\:\mathrm{dx}\: \\…

factorise-x-xy-2y-2x-

Question Number 141994 by jlewis last updated on 25/May/21 $$\mathrm{factorise}\:\:\mathrm{x}−\sqrt{\mathrm{xy}\:}\:+\sqrt{\mathrm{2y}}\:−\sqrt{\mathrm{2x}\:}\: \\ $$ Answered by Dwaipayan Shikari last updated on 25/May/21 $${x}−\sqrt{{xy}}+\sqrt{\mathrm{2}{y}}−\sqrt{\mathrm{2}{x}} \\ $$$$=\sqrt{{x}}\left(\sqrt{{x}}−\sqrt{{y}}\right)−\sqrt{\mathrm{2}}\left(\sqrt{{x}}−\sqrt{{y}}\right) \\ $$$$=\left(\sqrt{{x}}−\sqrt{{y}}\right)\left(\sqrt{{x}}−\sqrt{\mathrm{2}}\right)…

In-a-triangle-whose-sides-measure-5-cm-6-cm-and-7-cm-a-point-P-inside-the-triangle-is-2-cm-from-de-long-side-5-cm-and-3-cm-from-the-long-side-6-cm-What-is-the-distance-from-P-beside-7-cm-side-

Question Number 76455 by Maclaurin Stickker last updated on 27/Dec/19 $${In}\:{a}\:{triangle}\:{whose}\:{sides}\:{measure} \\ $$$$\mathrm{5}\:{cm},\:\mathrm{6}\:{cm},\:{and}\:\:\mathrm{7}\:{cm}\:{a}\:{point}\:{P},\:{inside} \\ $$$${the}\:{triangle}\:{is}\:\mathrm{2}\:{cm}\:{from}\:{de}\:{long}\:{side} \\ $$$$\mathrm{5}\:{cm}\:{and}\:\mathrm{3}\:{cm}\:{from}\:{the}\:{long}\:{side} \\ $$$$\mathrm{6}\:{cm}.\:{What}\:{is}\:{the}\:{distance}\:{from}\:{P} \\ $$$${beside}\:\mathrm{7}\:{cm}\:{side}? \\ $$ Commented by…

hence-show-that-i-1-cos4-sin4-tan2-ii-1-cos6-sin6-tan3-

Question Number 10917 by okhema last updated on 02/Mar/17 $${hence}\:{show}\:{that} \\ $$$$\left.{i}\right)\frac{\mathrm{1}−{cos}\mathrm{4}\theta}{{sin}\mathrm{4}\theta}={tan}\mathrm{2}\theta \\ $$$$\left.{ii}\right)\frac{\mathrm{1}−{cos}\mathrm{6}\theta}{{sin}\mathrm{6}\theta}={tan}\mathrm{3}\theta \\ $$ Answered by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{we}\:\mathrm{use}\:\:\:\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta\:\Rightarrow\:\mathrm{2sin}^{\mathrm{2}}…