Question Number 160792 by cortano last updated on 06/Dec/21 $$\:\:\:\int\:\frac{\mathrm{sec}\:\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{2sec}\:\mathrm{x}}}\:\sqrt{\frac{\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}}{\mathrm{cosec}\:\mathrm{x}+\mathrm{cot}\:\mathrm{x}}}\:\mathrm{dx}\:=? \\ $$ Answered by chhaythean last updated on 06/Dec/21 $$=\int\frac{\frac{\mathrm{1}}{\mathrm{cosx}}}{\:\sqrt{\frac{\mathrm{cosx}+\mathrm{2}}{\mathrm{cosx}}}}×\sqrt{\frac{\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{sinx}}}{\frac{\mathrm{1}+\mathrm{cosx}}{\mathrm{sinx}}}}\mathrm{dx} \\ $$$$=\int\frac{\frac{\mathrm{1}}{\mathrm{cosx}}}{\:\sqrt{\frac{\mathrm{cosx}+\mathrm{2}}{\mathrm{cosx}}}}×\sqrt{\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{1}+\mathrm{cosx}}}\mathrm{dx} \\ $$$$=\int\frac{\mathrm{sinx}}{\:\sqrt{\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{2cosx}}}×\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosx}}\mathrm{dx}…
Question Number 160767 by cortano last updated on 06/Dec/21 $$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=?\: \\ $$ Answered by MJS_new last updated on 06/Dec/21 $$\underset{\mathrm{0}}…
Question Number 95227 by bobhans last updated on 24/May/20 $$\underset{−\pi} {\overset{\pi} {\int}}\:\mid\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x}\:\mid\:\mathrm{dx}\:=?\: \\ $$ Answered by john santu last updated on 24/May/20 Commented by john…
Question Number 95221 by mathmax by abdo last updated on 24/May/20 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{2}+\mathrm{cos}\theta\right)\mathrm{d}\theta \\ $$ Answered by mathmax by abdo last updated on 24/May/20…
Question Number 95213 by mathmax by abdo last updated on 24/May/20 $$\mathrm{find}\:\int\:\frac{\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)}}{\:\sqrt{\mathrm{x}+\mathrm{3}}}\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 95208 by Ar Brandon last updated on 24/May/20 $$\mathrm{i}\backslash\:\int\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ii}\backslash\:\int\frac{\mathrm{cosx}}{\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{iii}\backslash\:\int\frac{\mathrm{1}}{\mathrm{lnx}}\mathrm{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{iv}\backslash\:\int\sqrt{\mathrm{1}−\mathrm{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{v}\backslash\:\int\sqrt{\mathrm{sinx}}\mathrm{dx}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{vi}\backslash\:\int\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{vii}\backslash\:\int\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\:\:\:\:\:\mathrm{viii}\backslash\:\int\mathrm{xtanxdx} \\ $$$$\mathrm{ix}\backslash\:\int\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\backslash\int\mathrm{e}^{\mathrm{x}^{\mathrm{2}}…
Question Number 95209 by Ar Brandon last updated on 24/May/20 $$\int\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 95198 by Fikret last updated on 23/May/20 $$\int\sqrt{\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}}{dx}=? \\ $$ Answered by MJS last updated on 23/May/20 $$\int\sqrt{\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}}{\mathrm{5}}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)}\right] \\ $$$$=−\mathrm{10}\int\frac{{t}^{\mathrm{2}}…
Question Number 160733 by cortano last updated on 05/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160734 by mnjuly1970 last updated on 05/Dec/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:#\:\mathrm{Advanced}\:\:\:\mathrm{Calculus}\:# \\ $$$$\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}^{\:} \:\left(\:\frac{\mathrm{1}}{\mathrm{1}−\:{x}}\:\:\right)}{{x}}\:\right)^{\:\mathrm{3}} {dx}\:\overset{?} {=}\:\mathrm{3}\:\left(\:\zeta\:\left(\mathrm{2}\:\right)\:+\:\zeta\:\left(\mathrm{3}\:\right)\right) \\ $$$$\:\:\:\:\:\:−−−−\:\:{solution}−−−− \\ $$$$\:\:\:\:\:\:\:\:\Phi\:\overset{\mathrm{I}.\mathrm{B}.\mathrm{P}} {=}\:\left[\:\frac{\:\mathrm{1}}{\mathrm{2}{x}^{\:\mathrm{2}} }\:{ln}^{\:\mathrm{3}}…