Question Number 62122 by aliesam last updated on 15/Jun/19 $$\int{e}^{{cos}\left({x}\right)} {sin}\left({sin}\left({x}\right)\right)\:{dx}\: \\ $$ Commented by MJS last updated on 15/Jun/19 $$\int\mathrm{e}^{\mathrm{cos}\:{x}} \mathrm{sin}\:\mathrm{sin}\:{x}\:{dx}=−\frac{\mathrm{i}}{\mathrm{2}}\int\left(\mathrm{e}^{\frac{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{2}}} −\mathrm{e}^{−\frac{\mathrm{e}^{\mathrm{i}{x}}…
Question Number 62121 by ajfour last updated on 16/Jun/19 $${A}\:{cube}\:{of}\:{unit}\:{edge}\:{length}\: \\ $$$${is}\:{held}\:{before}\:{a}\:{plane}.\:{Prove}\:{that} \\ $$$${the}\:{sum}\:{of}\:{the}\:{squares}\:{of}\:{the} \\ $$$${projected}\:{lengths}\:{of}\:{edges}\:{of}\:{the}\:{cube} \\ $$$${on}\:{the}\:{plane}\:\left({irrespective}\:{of}\right. \\ $$$$\left.{the}\:{orientation}\:{of}\:{the}\:{cube}\right)\:{is}\:\mathrm{8}. \\ $$ Answered by mr…
Question Number 127654 by Algoritm last updated on 31/Dec/20 Answered by luismej last updated on 31/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127652 by faysal last updated on 31/Dec/20 $${determine}\:{the}\:{coordinate}\:{of}\:{a}\:{point}\:{which}\:{is}\:{on}\:{the} \\ $$$$\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{12}=\mathrm{0}\:{and}\:{the}\:{distance}\:{of}\:{the}\:{point}\:{from}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$${is}\:\mathrm{10}\:{unit} \\ $$ Answered by liberty last updated on 31/Dec/20 $$\:\mathrm{let}\:\mathrm{the}\:\mathrm{point}\:\mathrm{A}\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{and}\:\mathrm{distance}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 127653 by faysal last updated on 31/Dec/20 $${A}\left(\mathrm{1},\mathrm{1}\right),\:{B}\left(\mathrm{3},\mathrm{4}\right),\:{C}\left(\mathrm{5},−\mathrm{2}\right) \\ $$$${on}\:{the}\:{paralel}\:{line}\:{of}\:{AB}\:,\:{what}\:{is}\:{the} \\ $$$${distance}\:{of}\:{AC}\:{from}\:\left(\mathrm{4},\mathrm{4}\right)\:{point} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127649 by mnjuly1970 last updated on 31/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:{calculate}:: \\ $$$$\:\:\:\:\:\emptyset=\int_{\mathrm{0}} ^{\:\infty} {arctan}\left(\frac{\mathrm{1}}{{x}}\right).{sin}\left({x}\right){dx}=? \\ $$$$ \\ $$ Answered by mindispower last updated…
Question Number 62112 by aliesam last updated on 15/Jun/19 Answered by MJS last updated on 15/Jun/19 $$\mathrm{sin}^{\mathrm{10}} \:{x}\:+\mathrm{cos}^{\mathrm{10}} \:{x}\:=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\frac{\mathrm{5}}{\mathrm{128}}\mathrm{cos}\:\mathrm{8}{x}\:+\frac{\mathrm{15}}{\mathrm{32}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{63}}{\mathrm{128}}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\mathrm{cos}\:\mathrm{8}{x}\:+\mathrm{12cos}\:\mathrm{4}{x}\:+\frac{\mathrm{43}}{\mathrm{9}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{t}…
Question Number 127644 by Dwaipayan Shikari last updated on 31/Dec/20 $$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}−\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{3}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{5}}{\mathrm{4}}−\frac{\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\mathrm{6}}{\mathrm{5}}−…}}}}}} \\ $$ Commented by Dwaipayan Shikari last updated on 31/Dec/20 $${a}_{\mathrm{0}} +{a}_{\mathrm{0}} {a}_{\mathrm{1}} +{a}_{\mathrm{0}}…
Question Number 62109 by sandhyavs last updated on 15/Jun/19 $${Given}\:{that} \\ $$$$\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}}\right)\mathrm{tan}\:{x}=\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right). \\ $$$${Then}\:{find}\:\:\:\mathrm{sin}\:\mathrm{4}{x}. \\ $$ Commented by sandhyavs last updated on 15/Jun/19 $${Also}\:{tell}\:{me}\:{the}\:{steps}.{Please}… \\…
Question Number 127641 by Study last updated on 31/Dec/20 Answered by Dwaipayan Shikari last updated on 31/Dec/20 $$\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…={e}^{−\zeta'\left(\mathrm{0}\right)} ={e}^{{log}\left(\sqrt{\mathrm{2}\pi}\right)} =\sqrt{\mathrm{2}\pi} \\ $$ Commented by MJS_new…