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Author: Tinku Tara

Question-200575

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-200569

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-200570

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-200549

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-200605

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…