Question Number 94123 by M±th+et+s last updated on 17/May/20 $${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{2}{n}} \left({x}\right)}{{x}^{\mathrm{2}} }{d}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{2}{n}−\mathrm{1}} \left({x}\right)}{{x}}{dx} \\ $$$$ \\ $$$$ \\ $$…
Question Number 94119 by i jagooll last updated on 17/May/20 $$\int\:\mathrm{cot}^{−\mathrm{1}} \left(\sqrt{\mathrm{x}}\right)\:\mathrm{dx}\: \\ $$ Commented by i jagooll last updated on 17/May/20 $${u}\:=\:\mathrm{cot}^{−\mathrm{1}} \left(\sqrt{{x}}\right)\:\Rightarrow\:{du}\:=\:\frac{{dx}}{\mathrm{2}\sqrt{{x}}\:\left(\mathrm{1}−{x}\right)} \\…
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Question Number 94098 by niroj last updated on 16/May/20 $$\:\:\mathrm{Integrate}: \\ $$$$\:\:\int\:\frac{\:\:\mathrm{dx}}{\mathrm{a}\:\mathrm{sin}\:\mathrm{x}+\:\mathrm{b}\:\mathrm{cos}\:\mathrm{x}} \\ $$ Commented by prakash jain last updated on 16/May/20 $$\mathrm{see}\:\mathrm{question}\:\mathrm{93098} \\ $$…
Question Number 94093 by MAB last updated on 16/May/20 $${evaluate}\:{the}\:{inequality}\:{for}\:{n}\geqslant\mathrm{2} \\ $$$$\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{n}}\right)\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}}}<\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{\pi}{\mathrm{2}}} \sqrt[{{n}}]{{sin}\left({t}\right)}{dt} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94084 by seedhamaieng@gmail.com last updated on 16/May/20 $$\int\sqrt{\mathrm{cot}{x}}{dx}\: \\ $$ Commented by Kunal12588 last updated on 16/May/20 $$=\int\sqrt{{tan}\left(\pi/\mathrm{2}−{x}\right)}\:{dx} \\ $$$$=−\int\sqrt{{tan}\:{t}}\:{dt}\:;\:{with}\:{t}={tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\int\sqrt{{tan}\:{x}}\:{dx}\:{is}\:{very}\:{popular}\:{integral},\:{its}\:{already} \\…
Question Number 28543 by abdo imad last updated on 26/Jan/18 $${find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} +{t}\:+{h}^{\mathrm{2}} \right)^{\mathrm{2}} \:+{h}^{\mathrm{2}} }\:\:\:. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 159612 by cortano last updated on 19/Nov/21 Answered by Ar Brandon last updated on 19/Nov/21 $${I}=\int\frac{\sqrt{\mathrm{2}+\sqrt[{\mathrm{3}}]{{x}}}}{\:\sqrt[{\mathrm{3}}]{{x}}}{dx},\:{x}={u}^{\mathrm{3}} \Rightarrow{dx}=\mathrm{3}{u}^{\mathrm{2}} {du} \\ $$$$\:\:\:=\mathrm{3}\int\frac{\sqrt{\mathrm{2}+{u}}}{{u}}\centerdot{u}^{\mathrm{2}} {du}=\mathrm{3}\int{u}\sqrt{\mathrm{2}+{u}}{du} \\ $$$$\:\:\:=\mathrm{3}\int\left({u}+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 28540 by abdo imad last updated on 26/Jan/18 $${find}\:\boldsymbol{{F}}\left(\:{e}^{−{ax}^{\mathrm{2}} } \right)\:\:\:{where}\:\:\boldsymbol{{F}}\:\:{mean}\:{fourier}\:{transform}. \\ $$$${a}>\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28541 by abdo imad last updated on 26/Jan/18 $${prove}\:{that}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{e}^{{ax}} −\mathrm{1}}{dx}=\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{p}^{\mathrm{2}} {a}^{\mathrm{2}} }\:\:\:\:\:{with}\:{a}>\mathrm{0} \\ $$ Terms of Service Privacy Policy…