Question Number 186170 by normans last updated on 01/Feb/23 $$ \\ $$$$\:\:\:\:\:\:\underset{−\mathrm{2}} {\overset{\mathrm{2}} {\int}}\:\:\:\:\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \:\left(\:\mathrm{2}\:−\:\boldsymbol{{cos}}\:\left(\boldsymbol{{x}}\right)\right)\:\:\:\:}{\mathrm{2}\:+\:\boldsymbol{{x}}^{\mathrm{2}} }\:\:\boldsymbol{{dx}} \\ $$$$ \\ $$ Answered by MJS_new last updated…
Question Number 186171 by normans last updated on 01/Feb/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{{so}}\:\boldsymbol{{easy}}\right] \\ $$$$\:\:\:\:\:\:\:\int\:\boldsymbol{{cos}}^{\mathrm{2}} \:\left(\mathrm{4}\boldsymbol{{x}}\right)\:+\:\boldsymbol{{sin}}^{\mathrm{4}} \:\left(\mathrm{2}\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}\:\:\: \\ $$$$ \\ $$ Answered by MJS_new last updated…
Question Number 120628 by TANMAY PANACEA last updated on 01/Nov/20 $${selective}\:{intregals} \\ $$ Commented by TANMAY PANACEA last updated on 01/Nov/20 $${thanks} \\ $$$$\int_{{a}} ^{{b}}…
Question Number 186152 by normans last updated on 01/Feb/23 $$ \\ $$$$\:\:\:\boldsymbol{{I}}=\:\:\underset{\mathrm{2}} {\overset{\boldsymbol{\pi}} {\int}}\:\:\:\frac{\boldsymbol{{cos}}^{\mathrm{2}} \:\left(\boldsymbol{{x}}\right)\:−\:\mathrm{1}\:}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\left(\boldsymbol{{x}}\right)\:−\:\boldsymbol{{tan}}\:\left(\boldsymbol{{x}}\right)}\:\:\boldsymbol{{dx}}\:\:\:\: \\ $$$$ \\ $$ Answered by MJS_new last updated on…
Question Number 55068 by gunawan last updated on 17/Feb/19 $$\mathrm{Calculate}\:\mathrm{value}\:\mathrm{of}\:\int_{{C}} {e}^{\frac{\mathrm{2}}{{z}}} \:{dz} \\ $$$$\mathrm{if}\:{C}\:\mathrm{is}\:\mathrm{unit}\:\mathrm{circle} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 55067 by gunawan last updated on 17/Feb/19 $$\mathrm{Known}\:\:{C}\:\mathrm{circle}\:\mathrm{centered}\:\mathrm{at}\:\mathrm{0}. \\ $$$$\mathrm{Find}\:\mathrm{value}\:\mathrm{than}\:\int_{{C}} \frac{{dz}}{\mathrm{1}−{z}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186132 by normans last updated on 01/Feb/23 Commented by MJS_new last updated on 01/Feb/23 $$\mathrm{I}\:\mathrm{will}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{and}\:\mathrm{your}\:\mathrm{earlier}\:\mathrm{integral} \\ $$$$\mathrm{immediately}\:\mathrm{after}\:\mathrm{I}\:\mathrm{finished}\:\mathrm{my}\:\mathrm{proof}\:\mathrm{of} \\ $$$$\mathrm{Riemann}'\mathrm{s}\:\mathrm{Hypothesis}. \\ $$$$\mathrm{see}\:\mathrm{you}\:\mathrm{later}\:\mathrm{alligator}! \\ $$…
Question Number 120599 by MagdiRagheb last updated on 01/Nov/20 $${Let}\:{x}\:=\:{u}^{\mathrm{6}} \:\:\:\:\:\:\:\:\:{dx}\:=\:\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$${I}\:=\:\int\frac{{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:×\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\frac{{u}^{\mathrm{8}} }{\mathrm{1}+\mathrm{2}{u}^{\mathrm{2}} +{u}^{\mathrm{4}} }\:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\left[\frac{−\mathrm{4}}{{u}^{\mathrm{2}}…
Question Number 55051 by maxmathsup by imad last updated on 16/Feb/19 $${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+…+{x}^{{n}} } \\ $$ Commented by maxmathsup by imad last…
Question Number 120582 by MagdiRagheb last updated on 01/Nov/20 $${Let}\:{u}={x}^{\frac{\mathrm{3}}{\mathrm{5}}} \:\:\:\:\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{5}{x}^{\frac{\mathrm{2}}{\mathrm{5}}} } \\ $$$${I}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\int\frac{{u}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du}\:=\:−\frac{\mathrm{5}}{\mathrm{6}}\:\int\frac{\mathrm{3}−\mathrm{2}{u}−\mathrm{3}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\:\int\left[\sqrt{\mathrm{3}−\mathrm{2}{u}}\:−\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right]\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\mathrm{3}+\mathrm{2}{u}\:+\:\mathrm{9}\right]+{c} \\…