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Question Number 200187 by sonukgindia last updated on 15/Nov/23 Answered by Mathspace last updated on 15/Nov/23 $${I}={Re}\left(\int_{−\infty} ^{+\infty} \frac{{e}^{{iax}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\right) \\ $$$${and}\:\int_{−\infty} ^{+\infty} \frac{{e}^{{iax}}…

Question Number 200122 by Blackpanther last updated on 14/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 200110 by sonukgindia last updated on 14/Nov/23 Answered by mr W last updated on 14/Nov/23 $${x}^{{x}^{{x}} } ={x}^{\mathrm{3}} \\ $$$$\Rightarrow{x}=\pm\mathrm{1} \\ $$$$\Rightarrow{x}^{{x}} =\mathrm{3}\Rightarrow{e}^{\mathrm{ln}\:{x}}…

Question Number 200060 by sonukgindia last updated on 13/Nov/23 Answered by cortano12 last updated on 13/Nov/23 $$\:\mathrm{L}=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\mathrm{2}{ab}−{a}\sqrt{{ab}}−{ab}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ab}−{a}\sqrt{{ab}}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\sqrt{{ab}}\:\left(\sqrt{{ab}}−{a}\right)}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)}\: \\…

Question Number 200075 by sonukgindia last updated on 13/Nov/23 Answered by ajfour last updated on 13/Nov/23 $${R}\mathrm{cos}\:\theta=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${R}={b}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{2}{b}\mathrm{cos}\:^{\mathrm{2}} \theta={a}+{b} \\ $$$$\frac{{a}}{{b}}=\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}…