Question Number 201091 by MrGHK last updated on 29/Nov/23 Commented by Frix last updated on 29/Nov/23 $$\mathrm{Look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}\:\mathrm{summands}: \\ $$$${i}=\mathrm{0}\:\rightarrow\:\mathrm{1} \\ $$$${i}=\mathrm{1}\:\rightarrow\:\frac{{n}}{\mathrm{2}{n}+\mathrm{4}} \\ $$$${i}=\mathrm{2}\:\rightarrow\:\frac{{n}^{\mathrm{2}} −{n}}{\mathrm{6}{n}+\mathrm{24}{n}+\mathrm{24}} \\…
Question Number 201035 by sonukgindia last updated on 28/Nov/23 Commented by mr W last updated on 28/Nov/23 $${the}\:{side}\:{length}\:{of}\:{the}\:{square}\:{can}\:{not} \\ $$$${be}\:{equal}\:{to}\:{the}\:{radius}\:{of}\:{the}\:{smaller} \\ $$$${semi}−{circles}! \\ $$ Commented…
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Question Number 201027 by sonukgindia last updated on 28/Nov/23 Answered by Atomist last updated on 28/Nov/23 $${sechx}=\frac{\mathrm{1}}{{coshx}}\: \\ $$$${coshx}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\int{sechxdx}=\int\frac{\mathrm{2}}{{e}^{{x}} +{e}^{−{x}} }{dx}=…
Question Number 200984 by sonukgindia last updated on 27/Nov/23 Answered by Calculusboy last updated on 29/Nov/23 $$\mathrm{2} \\ $$ Answered by witcher3 last updated on…
Question Number 200980 by sonukgindia last updated on 27/Nov/23 Answered by MM42 last updated on 27/Nov/23 $$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}}={y}\Rightarrow\:{x}={y}^{\mathrm{2}} −{y}\Rightarrow{dx}=\left(\mathrm{2}{y}−\mathrm{1}\right){dy} \\ $$$${y}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\:\:\:\:\:\Rightarrow\:\:\:{x}=\mathrm{0}\rightarrow{y}=\mathrm{1}\:\:\:\:;\:\:{x}=\mathrm{1}\rightarrow{y}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left.\Rightarrow\int_{\mathrm{1}} ^{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \:\:\frac{\mathrm{2}{y}−\mathrm{1}}{{y}}\:{dy}\:=\left(\mathrm{2}{y}−{lny}\right)\right]_{\mathrm{1}} ^{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}…
Question Number 200976 by Blackpanther last updated on 27/Nov/23 Answered by Mathspace last updated on 28/Nov/23 $$\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{cos}\left({n}\pi\right)}{{ln}\mathrm{3}}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{ln}\mathrm{3}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}}…
Question Number 200978 by sonukgindia last updated on 27/Nov/23 Answered by BaliramKumar last updated on 27/Nov/23 $$\mathrm{put}\:\:\:\mathrm{x}\:=\:\mathrm{tan}\theta\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\mathrm{dx}\:=\:\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \theta+\mathrm{1}}}\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\int\:\mathrm{sec}\theta\mathrm{d}\theta\:=\:\mathrm{ln}\left(\mathrm{sec}\theta\:+\:\mathrm{tan}\theta\right) \\…
Question Number 200979 by sonukgindia last updated on 27/Nov/23 Answered by Frix last updated on 27/Nov/23 $$\mathrm{Qu}\:\mathrm{200934} \\ $$ Terms of Service Privacy Policy Contact:…