Question Number 129958 by bemath last updated on 21/Jan/21 $$\:\mathrm{3}+\mathrm{cos}\:\left(\mathrm{6x}\right)\:=\:\mathrm{2}\:\left(\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{cos}\:\mathrm{4x}}\right)−\mathrm{4tan}\:^{\mathrm{2}} \left(\mathrm{4x}\right) \\ $$ Answered by MJS_new last updated on 21/Jan/21 $${s}=\mathrm{sin}\:{x} \\ $$$$… \\ $$$${s}^{\mathrm{14}}…
Question Number 64419 by turbo msup by abdo last updated on 17/Jul/19 $$\left.\mathrm{1}\right)\:{find}\:\int\:\:\frac{{dx}}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$ Commented by mathmax by…
Question Number 129955 by mohammad17 last updated on 21/Jan/21 Answered by Olaf last updated on 21/Jan/21 $$\mathrm{A}\:=\:\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{15}}&{\mathrm{4}}\\{\mathrm{16}}&{−\mathrm{2}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{17}}\\{\mathrm{2}}&{−\mathrm{14}}&{\mathrm{3}}&{\mathrm{2}}\end{bmatrix} \\ $$$$\mathrm{det}\left(\mathrm{A}\right)\:=\:−\mathrm{56460}\:\neq\:\mathrm{0} \\ $$$$\mathrm{A}_{\mathrm{1}} \:=\:\begin{bmatrix}{−\mathrm{2}}&{\mathrm{1}}&{−\mathrm{15}}&{\mathrm{4}}\\{\mathrm{2}}&{−\mathrm{2}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{9}}&{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{17}}\\{−\mathrm{8}}&{−\mathrm{14}}&{\mathrm{3}}&{\mathrm{2}}\end{bmatrix} \\ $$$$\mathrm{det}\left(\mathrm{A}_{\mathrm{1}} \right)\:=\:−\mathrm{14115}…
Question Number 64418 by turbo msup by abdo last updated on 17/Jul/19 $$\left.\mathrm{1}\right){find}\:\int\:\:\:\frac{{dx}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$ Commented by Prithwish sen…
Question Number 129951 by Algoritm last updated on 21/Jan/21 Commented by MJS_new last updated on 21/Jan/21 $$\mathrm{2}^{{x}/\mathrm{3}} −\mathrm{2}^{{x}} =\mathrm{2} \\ $$$${t}=\mathrm{2}^{{x}/\mathrm{3}} \\ $$$${t}−{t}^{\mathrm{3}} =\mathrm{2} \\…
Question Number 129949 by bemath last updated on 21/Jan/21 $$\mathrm{Given}\:\mathrm{8}\sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\:\right)\mathrm{dy}\:=\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}}}\:,\:\mathrm{x}>\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{y}\left(\mathrm{0}\right)=\sqrt{\mathrm{7}}\:.\:\mathrm{Find}\:\mathrm{y}\left(\mathrm{256}\right). \\ $$ Answered by liberty last updated on 21/Jan/21 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{8}\sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\right)\left(\sqrt{\mathrm{4}+\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}}\right)} \\ $$$$\:\mathrm{let}\:\sqrt{\mathrm{4}+\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\:}\:=\:\mathrm{z}\:;\:\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\:=\:\mathrm{z}^{\mathrm{2}} −\mathrm{4}…
Question Number 129946 by bemath last updated on 21/Jan/21 $$\:\int\:\frac{\mathrm{tan}\:\varphi+\mathrm{3}}{\mathrm{sin}\:\varphi}\:\mathrm{d}\varphi\:? \\ $$ Answered by liberty last updated on 21/Jan/21 $$\:\mathrm{T}\:=\:\int\:\left(\frac{\mathrm{1}}{\mathrm{cos}\:\varphi}\:+\:\frac{\mathrm{3}}{\mathrm{sin}\:\varphi}\right)\mathrm{d}\varphi \\ $$$$\:\mathrm{T}=\:\int\:\left(\mathrm{sec}\:\varphi+\mathrm{3csc}\:\varphi\right)\:\mathrm{d}\varphi\: \\ $$$$\:\mathrm{T}\:=\:\mathrm{ln}\:\mid\mathrm{sec}\:\varphi+\mathrm{tan}\:\varphi\mid−\mathrm{3ln}\:\mid\mathrm{cot}\:\varphi+\mathrm{csc}\:\varphi\mid\:+\:\mathrm{c} \\…
Question Number 64410 by mmkkmm000m last updated on 17/Jul/19 $$\int\mathrm{1}/\left(\mathrm{1}+{ysin}\theta\right){d}\theta \\ $$ Commented by mr W last updated on 17/Jul/19 $${see}\:{Q}\mathrm{64238}\:{sir}! \\ $$ Commented by…
Question Number 64406 by rajesh4661kumar@gamil.com last updated on 17/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${area}\:{of}\:\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\mid\begin{vmatrix}{{x}_{\mathrm{2}} −{x}_{\mathrm{1}} \:\:{y}_{\mathrm{2}} −{y}_{\mathrm{1}} }\\{{x}_{\mathrm{3}} −{x}_{\mathrm{1}} \:\:{y}_{\mathrm{3}} −{y}_{\mathrm{1}}…
Question Number 64404 by aliesam last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 $$\mathrm{x}+\mathrm{1}\:=\:\sqrt{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\left(\mathrm{x}+\mathrm{2}\right)}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }} \\ $$$$=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }}} \\ $$$$=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{4}\right)}}\:}…