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Question Number 176540 by yaslm last updated on 20/Sep/22 Answered by Peace last updated on 21/Sep/22 $$\mathcal{F}\left({x}\right)=\int\frac{{dx}}{{cos}^{\mathrm{2}} \left({x}\right)\sqrt{{cos}\left({x}\right)}}={tg}\left({x}\right)\frac{\mathrm{1}}{\:\sqrt{{cos}\left({x}\right)}}−\int{tg}\left({x}\right).\frac{{sin}\left({x}\right)}{\mathrm{2}\sqrt{{cos}\left({x}\right)}{cos}\left({x}\right)} \\ $$$$−\int\frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)\sqrt{{cos}\left({x}\right)}}{dx} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\mathcal{F}\left({x}\right)=\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)\sqrt{{cos}\left({x}\right)}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{{cos}\left({x}\right)}} \\…

Question Number 45433 by mondodotto@gmail.com last updated on 12/Oct/18 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{differnt}}\:\boldsymbol{\mathrm{ways}}\: \\ $$$$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{cricket}}\:\boldsymbol{\mathrm{team}}\:\boldsymbol{\mathrm{consisting}}\:\boldsymbol{\mathrm{of}} \\ $$$$\mathrm{11}\boldsymbol{\mathrm{people}},\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{choosen}} \\ $$$$\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{group}}\:\boldsymbol{\mathrm{of}}\:\mathrm{16} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18…

Question Number 110964 by mathdave last updated on 01/Sep/20 $${solve}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$ Answered by mathdave last updated on…

Question Number 110951 by ZiYangLee last updated on 01/Sep/20 Answered by $@y@m last updated on 01/Sep/20 $${Let}\:{y}={kx}\:\left(\mathrm{0}<{k}<\mathrm{1}\right) \\ $$$${Let}\:{z}={x}+\frac{\mathrm{1}}{{y}\left({x}−{y}\right)} \\ $$$$\Rightarrow\:{z}={x}+\frac{\mathrm{1}}{{kx}\left({x}−{kx}\right)} \\ $$$$\:\Rightarrow{z}\:={x}+\frac{\mathrm{1}}{{k}\left(\mathrm{1}−{k}\right){x}^{\mathrm{2}} }\:…..\left(\mathrm{1}\right) \\…

Question Number 110944 by bemath last updated on 01/Sep/20 $$\:\:\:\blacksquare\sqrt{\mathrm{bemath}}\bigstar \\ $$$$\left(\mathrm{1}\right)\mathrm{If}\:\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{b}}\:=\:\mathrm{20}\:,\:\mathrm{a},\mathrm{b}\in\mathbb{R}\:,\:\mathrm{find}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{a}−\mathrm{5b}\:? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{3}\sqrt{\mathrm{x}}−\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} −\mathrm{16}}\:? \\ $$$$\left(\mathrm{3}\right)\int\:\frac{\mathrm{tan}\:\left(\mathrm{ln}\:\mathrm{x}\right)\:\mathrm{tan}\:\left(\mathrm{ln}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}{\mathrm{x}}\:\mathrm{dx} \\ $$$$\left(\mathrm{4}\right)\frac{\left(\sqrt{\mathrm{3x}−\mathrm{7}}\right)^{\mathrm{2}} −\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}−\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{3}} \\…