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0-pi-2-tanx-1-7-dx-

Question Number 122176 by Dwaipayan Shikari last updated on 14/Nov/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt[{\mathrm{7}}]{{tanx}}\:{dx} \\ $$ Commented by Dwaipayan Shikari last updated on 14/Nov/20 $${I}\:{have}\:{found}\:\frac{\pi}{\mathrm{2}}{cosec}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\…

Question-187662

Question Number 187662 by Jonas007 last updated on 20/Feb/23 Answered by Ar Brandon last updated on 20/Feb/23 $${I}=\int\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}{x}+\mathrm{cos}{x}\right)^{\mathrm{2}} }{dx}=\int\frac{{x}\mathrm{cos}{x}}{\left({x}\mathrm{sin}{x}+\mathrm{cos}{x}\right)^{\mathrm{2}} }\centerdot\frac{{x}}{\mathrm{cos}{x}}{dx} \\ $$$$\begin{cases}{\mathrm{u}\left({x}\right)=\frac{{x}}{\mathrm{cos}{x}}}\\{\mathrm{v}'\left({x}\right)=\frac{{x}\mathrm{cos}{x}}{\left({x}\mathrm{sin}{x}+\mathrm{cos}{x}\right)^{\mathrm{2}} }}\end{cases}\:\Rightarrow\begin{cases}{\mathrm{u}'\left({x}\right)=\frac{\mathrm{cos}{x}+{x}\mathrm{sin}{x}}{\mathrm{cos}^{\mathrm{2}} {x}}}\\{\mathrm{v}\left({x}\right)=−\frac{\mathrm{1}}{{x}\mathrm{sin}{x}+\mathrm{cos}{x}}}\end{cases}…

It-is-my-kind-request-to-those-who-post-questions-pls-go-through-the-details-of-answer-and-give-feed-back-pls-activate-yourselves-to-pay-your-attention-in-the-details-of-answer-do-not-become-

Question Number 56565 by tanmay.chaudhury50@gmail.com last updated on 18/Mar/19 $${It}\:{is}\:{my}\:{kind}\:{request}\:{to}\:{those}\:{who}\:{post}\:{questions} \\ $$$$…{pls}\:{go}\:{through}\:{the}\:{details}\:{of}\:{answer}…{and}\:{give} \\ $$$${feed}\:{back}…{pls}\:{activate}\:{yourselves}\:{to}\:{pay}\:{your} \\ $$$${attention}\:{in}\:{the}\:{details}\:{of}\:{answer}…{do}\:{not}\:{become} \\ $$$${self}\:{satisfied}\:{by}\:{getting}\:{your}\:{desired}\:{results}.. \\ $$$${unfurl}\:{your}\:{mind}\:{and}\:{act}\:{in}\:{such}\:{away}\:{that} \\ $$$${we}\:{get}\:{a}\:{tip}\:{of}\:{iceberg}\:{of}\:{your}\:{satisfation}.{Tanmay} \\ $$ Commented…

Question-187639

Question Number 187639 by LowLevelLump last updated on 19/Feb/23 Commented by a.lgnaoui last updated on 19/Feb/23 $${The}\:{question}\:{is}\:{like} \\ $$$${v}_{{n}+\mathrm{1}} −{v}_{{n}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\frac{{S}_{{n}}…