Question Number 43999 by maxmathsup by imad last updated on 19/Sep/18 $${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{x}}{\:\sqrt{\mathrm{1}−{cosx}}}{dx}. \\ $$ Commented by maxmathsup by imad last updated on 20/Sep/18…
Question Number 43991 by Tawa1 last updated on 19/Sep/18 $$\int\:\frac{\mathrm{1}}{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\:\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:\:\:\mathrm{dx} \\ $$ Answered by Joel578 last updated on 19/Sep/18 $${u}\:=\:\:{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \:\:\rightarrow\:\:{u}^{\mathrm{6}} \:=\:{x} \\…
Question Number 43987 by Necxx last updated on 19/Sep/18 Commented by MJS last updated on 19/Sep/18 $$???\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}??? \\ $$ Commented by Necxx last…
Question Number 43981 by Necxx last updated on 19/Sep/18 $$\int\sqrt{\mathrm{sin}\:{x}}\:{dx} \\ $$ Commented by MJS last updated on 19/Sep/18 $$\mathrm{see}\:\mathrm{question}\:\mathrm{42945} \\ $$$$\mathrm{you}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{scroll}\:\mathrm{down}\:\mathrm{a}\:\mathrm{while},\:\mathrm{Sir}\:\mathrm{Tanmay} \\ $$$$\mathrm{has}\:\mathrm{found}\:\mathrm{a}\:\mathrm{way}\:\mathrm{and}\:\mathrm{I}\:\mathrm{also}\:\mathrm{added}\:\mathrm{some}\:\mathrm{useful} \\…
Question Number 175051 by Mathspace last updated on 17/Aug/22 $${find}\:{the}\:{value}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{arctanx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$ Commented by mokys last updated on 20/Aug/22…
Question Number 109509 by bemath last updated on 24/Aug/20 $$\:\:\frac{{bemath}}{\underset{{i}={cooll}} {\overset{{nice}} {\sum}}\left({joss}\right)_{{i}} }\: \\ $$$$ \\ $$$$\int\:\frac{{x}^{\mathrm{2}} \:{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$ Answered by Dwaipayan Shikari…
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Question Number 175042 by peter frank last updated on 17/Aug/22 Answered by Frix last updated on 18/Aug/22 $$\int\frac{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{x}\:+{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}}{{a}^{\mathrm{4}} \mathrm{sin}^{\mathrm{4}} \:{x}\:+{b}^{\mathrm{4}} \mathrm{cos}^{\mathrm{4}}…
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