Question Number 112847 by bemath last updated on 10/Sep/20 $$\:\int\:\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$ Answered by bobhans last updated on 10/Sep/20 $$\mathrm{I}\:=\:\int\:\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{x}−\mathrm{2sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\:\mathrm{dx} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{x}\:\mathrm{cosec}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:\mathrm{dx}−\int\:\mathrm{cot}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{dx} \\…
Question Number 47311 by Meritguide1234 last updated on 08/Nov/18 Commented by maxmathsup by imad last updated on 09/Nov/18 $${changement}\:{x}+\mathrm{2}={t}\:{give}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{5}}{dx}\:=\int_{\mathrm{2}} ^{+\infty}…
Question Number 47295 by maxmathsup by imad last updated on 07/Nov/18 $${calculate}\:{f}\left(\alpha\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:{cos}\alpha\:+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left(\alpha\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{sin}\alpha}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\:{cos}\alpha+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{f}^{\left({n}\right)} \left(\alpha\right)\:{with}\:{n}\:{integr}\:{natural}\:. \\…
Question Number 178341 by haladu last updated on 15/Oct/22 Answered by mahdipoor last updated on 15/Oct/22 $${log}_{{y}} {x}=\frac{{lnx}}{{lny}}\Rightarrow{log}_{{b}} {a}+{log}_{{c}} {a}+{log}_{{d}} {a}= \\ $$$${ln}\left({a}\right).\left(\frac{\mathrm{1}}{{ln}\left({b}\right)}+\frac{\mathrm{1}}{{ln}\left({c}\right)}+\frac{\mathrm{1}}{{ln}\left({d}\right)}\right)={q}.{ln}\left({a}\right) \\ $$$$\int\:\frac{{sin}\left({q}.{lna}\right)}{{a}^{\mathrm{2}}…
Question Number 47259 by Ac Kesharwani last updated on 07/Nov/18 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}+\mathrm{1}}\right)}{\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)}\boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 112797 by mnjuly1970 last updated on 09/Sep/20 $$\:\:\:\:\:\:\:\:\:….{calculus}… \\ $$$$\:\:\:{evaluate} \\ $$$$ \\ $$$${i}:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}\sqrt{\:{tan}\left({x}\right)}\:{dx}=\:???\: \\ $$$${ii}:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}\:=???\: \\…
Question Number 112792 by mnjuly1970 last updated on 09/Sep/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 47248 by maxmathsup by imad last updated on 06/Nov/18 $${calculate}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{{ln}\left({x}+\mathrm{2}\right)}{\left({x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 47182 by maxmathsup by imad last updated on 05/Nov/18 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} \sqrt{\mathrm{1}−\sqrt{{x}}}{dx}\: \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18 Commented…
Question Number 178246 by mnjuly1970 last updated on 14/Oct/22 Answered by CElcedricjunior last updated on 14/Oct/22 $$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{64}}\left(\boldsymbol{{a}\zeta}\left(\mathrm{2}\right)+{b}\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\infty}…