Question Number 200575 by sonukgindia last updated on 20/Nov/23 Answered by AST last updated on 20/Nov/23 $$\frac{{ra}}{\mathrm{2}}=\frac{{bx}}{\mathrm{2}}\Rightarrow{ra}={bx};{x}=\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} };{b}={r}−{a} \\ $$$${ra}={bx}\Rightarrow{ra}=\left({r}−{a}\right)\left(\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\mathrm{2}{r}={a}\underset{−}…
Question Number 200568 by Rupesh123 last updated on 20/Nov/23 Commented by AST last updated on 20/Nov/23 $${Cannot}\:{be}\:{uniquely}\:{determined}. \\ $$ Commented by Frix last updated on…
Question Number 200569 by Rupesh123 last updated on 20/Nov/23 Answered by MM42 last updated on 20/Nov/23 $$\bigstar\:\:{e}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}!}\:\Rightarrow\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}!}={e}−\mathrm{2} \\ $$$$\sqrt{{x}\sqrt[{\mathrm{3}}]{{x}\sqrt[{\mathrm{4}}]{{x}\sqrt[{\mathrm{5}}]{\sqrt{{x}…}}}}}={x}^{\frac{\mathrm{1}}{\mathrm{2}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{6}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{24}}}…
Question Number 200570 by Rupesh123 last updated on 20/Nov/23 Answered by witcher3 last updated on 20/Nov/23 $$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}}=\mathrm{Im}\left\{\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{e}^{\mathrm{in}} }{\mathrm{n}}\right\} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\mathrm{e}^{\mathrm{in}} =\frac{\mathrm{e}^{\mathrm{i}}…
Question Number 200565 by hardmath last updated on 20/Nov/23 $$\mathrm{sin}^{\mathrm{10}} \mathrm{x}\:\:+\:\:\mathrm{cos}^{\mathrm{10}} \mathrm{x}\:\:=\:\:\frac{\mathrm{29}}{\mathrm{16}}\:\mathrm{cos}^{\mathrm{4}} \mathrm{2x} \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}\:=\:? \\ $$ Commented by Harnada last updated on 21/Nov/23 $${Q}\mathrm{200565}…
Question Number 200553 by cortano12 last updated on 20/Nov/23 $$\:\:\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\:\sqrt{\mathrm{1}+\mathrm{x}}}\:\mathrm{dx}\:=?\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200549 by sonukgindia last updated on 20/Nov/23 Answered by witcher3 last updated on 20/Nov/23 $$\mathrm{I}_{\mathrm{12}} =\mathrm{0};\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{I}_{\mathrm{11}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\int_{\mathrm{1}}…
Question Number 200604 by Calculusboy last updated on 20/Nov/23 Commented by Frix last updated on 21/Nov/23 $$\mathrm{Does}\:\mathrm{not}\:\mathrm{converge}. \\ $$ Commented by Calculusboy last updated on…
Question Number 200605 by Calculusboy last updated on 20/Nov/23 Commented by Frix last updated on 21/Nov/23 $$\mathrm{Simply}\:\mathrm{integrate}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{ln}\:\mid\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\mid\:−\frac{{x}}{\mathrm{sin}\:{x}}+{C} \\ $$ Commented by Calculusboy last…
Question Number 200606 by Calculusboy last updated on 20/Nov/23 Answered by Frix last updated on 20/Nov/23 $$\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}\:\overset{{t}={x}−\frac{\pi}{\mathrm{4}}} {=} \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos}\:{t}\:\mathrm{sin}\:{t}}{\mathrm{2}}−\frac{\mathrm{sin}^{\mathrm{2}} \:{t}}{\mathrm{2}}−\frac{\mathrm{tan}\:{t}}{\mathrm{4}}\right){dt}= \\ $$$$=\frac{{t}}{\mathrm{4}}−\frac{\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{4}}−\frac{{t}+\mathrm{cos}\:{t}\:\mathrm{sin}\:{t}}{\mathrm{4}}+\frac{\mathrm{ln}\:\mathrm{cos}\:{t}}{\mathrm{4}}=…