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0-pi-2-1-9-cos-x-12-sin-x-dx-

Question Number 65760 by mmkkmm000m last updated on 03/Aug/19 $$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{1}}{\mathrm{9}\:\mathrm{cos}\:{x}+\mathrm{12}\:\mathrm{sin}\:{x}}\:{dx}\:= \\ $$ Commented by mathmax by abdo last updated on 04/Aug/19 $${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…

Question-65753

Question Number 65753 by Masumsiddiqui399@gmail.com last updated on 03/Aug/19 Answered by mr W last updated on 04/Aug/19 $$\mathrm{1}\:\:\:^{\left.\ast\right)} \\ $$$$\sqrt{\mathrm{1}+\mathrm{1}} \\ $$$$\sqrt{\sqrt{\mathrm{1}+\mathrm{1}}+\mathrm{1}} \\ $$$$\sqrt{\sqrt{\sqrt{\mathrm{1}+\mathrm{1}}+\mathrm{1}}+\mathrm{1}} \\…

Question-131288

Question Number 131288 by mohammad17 last updated on 03/Feb/21 Answered by liberty last updated on 11/Feb/21 $$\mathrm{L}=\int\:\sqrt{\frac{\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} }{\mathrm{x}^{\mathrm{8}} }}\:\mathrm{dx}\:=\:\int\:\mathrm{x}^{−\mathrm{4}} \:\sqrt{\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} }\:\mathrm{dx} \\ $$$$\mathrm{change}\:\mathrm{of}\:\mathrm{variable}\::\:\sqrt{\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} }\:=\:\mathrm{h}\:\mathrm{or}\:\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} =\mathrm{h}^{\mathrm{2}}…

0-1-e-x-ln-x-dx-

Question Number 131283 by SEKRET last updated on 03/Feb/21 $$\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\mathrm{1}}{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}\:\boldsymbol{\mathrm{dx}}=? \\ $$$$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…

Question-65749

Question Number 65749 by bshahid010@gmail.com last updated on 03/Aug/19 Answered by Tanmay chaudhury last updated on 03/Aug/19 $${sin}^{\mathrm{2}} {x}+{sin}^{\mathrm{4}} {x}+{sin}^{\mathrm{6}} {x}+…\infty \\ $$$$=\frac{{sin}^{\mathrm{2}} {x}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}}…

x-2-8xy-17y-2-0-

Question Number 202 by 02@>@0 last updated on 25/Jan/15 $${x}^{\mathrm{2}} −\mathrm{8}{xy}+\mathrm{17}{y}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$ Answered by prakash jain last updated on 16/Dec/14 $${x}^{\mathrm{2}} −\mathrm{8}{xy}+\mathrm{17}{y}^{\mathrm{2}} \\…

7-20-9-20-

Question Number 203 by 02@>@0 last updated on 25/Jan/15 $$\frac{\mathrm{7}}{\mathrm{20}}+\frac{\mathrm{9}}{\mathrm{20}} \\ $$ Answered by 123456 last updated on 15/Dec/14 $$\frac{\mathrm{16}}{\mathrm{20}}=\frac{\mathrm{8}}{\mathrm{10}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$ Terms of Service…

Important-Information-If-you-are-a-student-and-not-able-to-pay-afford-premium-version-please-send-us-an-email-with-your-forum-user-id-to-get-free-premium-access-for-1-year-Please-make-sure-you-inclu

Question Number 131259 by Tinku Tara last updated on 03/Feb/21 $$\mathrm{Important}\:\mathrm{Information}: \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{are}\:\mathrm{a}\:\mathrm{student}\:\mathrm{and}\:\mathrm{not}\:\mathrm{able}\:\mathrm{to} \\ $$$$\mathrm{pay}/\mathrm{afford}\:\mathrm{premium}\:\mathrm{version}\:\mathrm{please} \\ $$$$\mathrm{send}\:\mathrm{us}\:\mathrm{an}\:\mathrm{email}\:\mathrm{with}\:\mathrm{your}\:\mathrm{forum} \\ $$$$\mathrm{user}\:\mathrm{id}\:\mathrm{to}\:\mathrm{get}\:\mathrm{free}\:\mathrm{premium}\:\mathrm{access} \\ $$$$\mathrm{for}\:\mathrm{1}\:\mathrm{year}. \\ $$$$\mathrm{Please}\:\mathrm{make}\:\mathrm{sure}\:\mathrm{you}\:\mathrm{include}\:\mathrm{your} \\ $$$$\mathrm{user}\:\mathrm{id}\:\mathrm{used}\:\mathrm{in}\:\mathrm{Q\&A}\:\mathrm{Forum}.…