Question Number 148798 by abdurehime last updated on 31/Jul/21 Answered by puissant last updated on 31/Jul/21 $$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\int\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}+\int\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\sqrt{\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)}}\mathrm{dx}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\sqrt{\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)}}\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\left(\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}+\sqrt{\mathrm{cotan}\left(\mathrm{x}\right)}\right)\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\left(\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}}\right)\mathrm{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{tan}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}}\mathrm{dx}…
Question Number 148785 by Sozan last updated on 31/Jul/21 $${find}\:{the}\:{resideu}\:{f}\left({z}\right)={z}^{−\mathrm{3}} {csc}\left({z}^{\mathrm{2}} \right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148783 by 0731619 last updated on 31/Jul/21 Commented by hknkrc46 last updated on 31/Jul/21 $$\left(\mathrm{1}\right)\:\underset{\boldsymbol{{h}}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{f}}\left(\boldsymbol{{x}}\:+\:\boldsymbol{{h}}\right)\:−\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{h}}}\: \\ $$$$=\:\underset{\boldsymbol{{h}}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{cos}\:\left(\boldsymbol{{x}}\:+\:\boldsymbol{{h}}\right)\:+\:\mathrm{cos}\:\boldsymbol{{x}}}{\boldsymbol{{h}}} \\ $$$$=\:\underset{\boldsymbol{{h}}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\boldsymbol{{x}}\:−\:\mathrm{cos}\:\boldsymbol{{x}}\:\centerdot\:\mathrm{cos}\:\boldsymbol{{h}}\:+\:\mathrm{sin}\:\boldsymbol{{x}}\:\centerdot\:\mathrm{sin}\:\boldsymbol{{h}}}{\boldsymbol{{h}}} \\…
Question Number 17675 by mondodotto@gmail.com last updated on 09/Jul/17 Answered by Tinkutara last updated on 10/Jul/17 $$\mathrm{Let}\:{a}^{{y}} \:=\:{x}\:\left(\boldsymbol{{i}}\right) \\ $$$$\mathrm{Taking}\:\mathrm{log}\:\mathrm{on}\:\mathrm{both}\:\mathrm{sides},\:\mathrm{log}_{{a}} \:{x}\:=\:{y} \\ $$$$\mathrm{Substituting}\:\mathrm{this}\:\mathrm{value}\:\mathrm{of}\:{y}\:\mathrm{in}\:\left(\boldsymbol{{i}}\right),\:\mathrm{we} \\ $$$$\mathrm{get}\:\boldsymbol{{a}}^{\boldsymbol{\mathrm{log}}_{\boldsymbol{{a}}}…
Question Number 148724 by Sozan last updated on 30/Jul/21 $${find}\:{laurent}\:{series}\:{f}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −{z}+\mathrm{1}}\:\:,\mathrm{0}<\mid{z}−\mathrm{1}\mid<\mathrm{1} \\ $$ Answered by mathmax by abdo last updated on 30/Jul/21 $$\mathrm{the}\:\mathrm{way}\:\mathrm{of}\:\mathrm{this}\:\mathrm{kind}\:\mathrm{is}\:\mathrm{to}\:\mathrm{use}\:\mathrm{changement}\:\mathrm{z}−\mathrm{1}=\mathrm{y}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\varphi\left(\mathrm{y}\right)=\frac{\mathrm{1}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 148733 by Rustambek last updated on 30/Jul/21 $${tg}\left(\alpha\right)+{tg}\left(\beta\right)=\mathrm{4} \\ $$$${cos}\left(\alpha\right)+{cos}\left(\beta\right)=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${tg}\left(\alpha+\beta\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148723 by Rustambek last updated on 30/Jul/21 $${tgx}+{tgy}=\mathrm{4}\:\:\:{cosx}+{cosy}=\frac{\mathrm{1}}{\mathrm{5}}\:\:\: \\ $$$${tg}\left({x}+{y}\right)=? \\ $$ Commented by Rustambek last updated on 30/Jul/21 $${please} \\ $$ Terms…
Question Number 148711 by Rankut last updated on 30/Jul/21 $${if}\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{4},\:{and}\:{x}−\frac{\mathrm{1}}{{x}}=\mathrm{3} \\ $$$${then}\:{prove}\:{that}\:\mathrm{4}=\mathrm{5} \\ $$ Answered by liberty last updated on 30/Jul/21 $$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{16}\Rightarrow\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 17600 by mondodotto@gmail.com last updated on 08/Jul/17 Answered by alex041103 last updated on 08/Jul/17 $$\mathrm{We}\:\mathrm{use}\:\mathrm{the}\:\mathrm{following}\:\mathrm{trig}\:\mathrm{sub}: \\ $$$${x}={tan}\left(\theta\right)\Rightarrow{dx}={sec}^{\mathrm{2}} \left(\theta\right){d}\theta \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{1}+{tan}^{\mathrm{2}} \theta={sec}^{\mathrm{2}} \theta \\…
Question Number 148645 by metamorfose last updated on 29/Jul/21 $${u}_{{n}+\mathrm{3}} =\frac{{u}_{{n}+\mathrm{2}} +{u}_{{n}+\mathrm{1}} +{u}_{{n}} }{\mathrm{3}}\:,\:\forall{n}\in{IN} \\ $$$${find}\:{u}_{{n}} \: \\ $$ Answered by Olaf_Thorendsen last updated on…