Question Number 97239 by john santu last updated on 07/Jun/20 $$\int\:\frac{\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}}{\:\sqrt{\mathrm{tan}\:{x}}}\:?\: \\ $$ Commented by MJS last updated on 07/Jun/20 $${t}=\sqrt{\mathrm{tan}\:{x}}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{2}\int\sqrt{{t}^{\mathrm{4}} +\mathrm{1}}{dt}\:\mathrm{and}\:\mathrm{we}\:\mathrm{cannot} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{using}\:\mathrm{elementary}\:\mathrm{calculus}…
Question Number 97235 by bobhans last updated on 07/Jun/20 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{ln}\left(\mathrm{x}\right)\:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:\mathrm{dx}\:?\: \\ $$ Answered by abdomathmax last updated on 07/Jun/20 $$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{lnxln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{for}\:\mid\mathrm{x}\mid<\mathrm{1} \\…
Question Number 97218 by bobhans last updated on 07/Jun/20 $$\underset{−\mathrm{2}} {\overset{\mathrm{3}} {\int}}\:\mid{x}−\mathrm{2}\mid\:\lfloor\:\frac{{x}}{\mathrm{2}}\:\rfloor\:\mathrm{sgn}\:\left({x}−\mathrm{1}\right)\:{dx}\: \\ $$ Answered by john santu last updated on 07/Jun/20 $$\mathrm{sgn}\left({x}−\mathrm{1}\right)=\mathrm{1}\:,\:\mathrm{when}\:{x}>\mathrm{1} \\ $$$$\mathrm{sgn}\left({x}−\mathrm{1}\right)=−\mathrm{1},\:\mathrm{when}\:{x}<\mathrm{1}…
Question Number 162702 by amin96 last updated on 31/Dec/21 $$\int\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{tg}}\left(\boldsymbol{{x}}\right)\boldsymbol{{ln}}\mid\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\mid\boldsymbol{{dx}}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 97153 by eidmarie last updated on 06/Jun/20 Answered by MJS last updated on 06/Jun/20 $$−\int\frac{{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{13}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{3}\right)}{dx}= \\ $$$$=−\int{dx}+\frac{\mathrm{13}}{\mathrm{5}}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{8}{x}−\mathrm{41}}{{x}^{\mathrm{2}} −{x}+\mathrm{3}}{dx}= \\ $$$$=−{x}+\frac{\mathrm{13}}{\mathrm{5}}\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:−\frac{\mathrm{4}}{\mathrm{5}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 97132 by student work last updated on 06/Jun/20 $$\mathrm{is}\:\mathrm{the}\:\mathrm{formulla}\:\mathrm{of}\:\mathrm{sin}\:^{\mathrm{3}} \left(\frac{\pi}{\mathrm{2}}+\mathrm{x}\right)=\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:\:\:\:\mathrm{correct}? \\ $$ Commented by student work last updated on 06/Jun/20 $$\mathrm{ans}\:\mathrm{me} \\…
Question Number 97130 by Quadri last updated on 06/Jun/20 Answered by Sourav mridha last updated on 06/Jun/20 $$=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{ln}}\left(\mathrm{2}\right)}\int\frac{\boldsymbol{{d}}\left(\mathrm{1}+\mathrm{2}^{\boldsymbol{{x}}^{\mathrm{2}} } \right)}{\left(\mathrm{1}+\mathrm{2}^{\boldsymbol{{x}}^{\mathrm{2}} } \right)}=\frac{\mathrm{1}}{\boldsymbol{{ln}}\left(\mathrm{4}\right)}.\boldsymbol{{ln}}\left(\mathrm{1}+\mathrm{2}^{\boldsymbol{{x}}^{\mathrm{2}} } \right)+\boldsymbol{{c}} \\…
Question Number 97109 by student work last updated on 06/Jun/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{3}} ×}\mathrm{dx}=? \\ $$ Answered by Sourav mridha last updated on…
Question Number 97091 by Mathudent last updated on 06/Jun/20 $${solve}\:\int{x}^{{x}+\mathrm{1}} {dx}\:. \\ $$ Answered by Sourav mridha last updated on 06/Jun/20 $$\int\boldsymbol{{e}}^{\left(\boldsymbol{{x}}+\mathrm{1}\right).\boldsymbol{{l}}\mathrm{n}\left(\boldsymbol{{x}}\right)} \boldsymbol{{dx}} \\ $$$$=\int\underset{\boldsymbol{{n}}=\mathrm{0}}…
Question Number 97089 by Mathudent last updated on 06/Jun/20 $${solve}\:\:\int{x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right){dx}\:. \\ $$ Answered by M±th+et+s last updated on 06/Jun/20 $${let}\:{x}^{{x}} ={u}\:\:\:\:{du}={x}^{{x}} \left(\mathrm{1}+{ln}\left({x}\right)\right){dx} \\ $$$$\int{du}={u}+{c}…