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Author: Tinku Tara

Reponse-a-l-exercice-N8-Reponses-par-ordre-1-2-3-4-5-6-imsge-1-imsge-2-image-3-imsge-5-imsge-4-imsge-6-

Question Number 216454 by a.lgnaoui last updated on 08/Feb/25 $$\mathrm{Reponse}\:\mathrm{a}\:\:\mathrm{l}\:\mathrm{exercice}\:\:\mathrm{N8}: \\ $$$$\mathrm{Reponses}\:\mathrm{par}\:\mathrm{ordre}:\left(\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right) \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{1}\: \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{2} \\ $$$$\boldsymbol{\mathrm{image}}\:\mathrm{3} \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{5} \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{4} \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{6} \\…

1-1-1-x-1-x-1-x-ln-2x-2-2x-1-2x-2-2x-1-dx-

Question Number 216408 by MrGaster last updated on 07/Feb/25 $$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\mathrm{ln}\left(\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\right){dx} \\ $$ Commented by MrGaster last updated on 07/Feb/25 $${I}=\int_{−\mathrm{1}} ^{+\mathrm{1}}…

If-asin-bcos-acosec-bsec-then-prove-that-each-term-is-equal-to-a-2-3-b-2-3-a-2-3-b-2-3-

Question Number 216421 by MATHEMATICSAM last updated on 07/Feb/25 $$\mathrm{If}\:{a}\mathrm{sin}\theta\:+\:{b}\mathrm{cos}\theta\:=\:{a}\mathrm{cosec}\theta\:+\:{b}\mathrm{sec}\theta\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{each}\:\mathrm{term}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left({a}^{\frac{\mathrm{2}}{\mathrm{3}}} \:−\:{b}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)\sqrt{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:{b}^{\frac{\mathrm{2}}{\mathrm{3}}} }. \\ $$ Answered by mr W last…

xe-x-x-1-2-dx-

Question Number 216372 by glory86 last updated on 06/Feb/25 $$\int\frac{{xe}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$ Answered by MrGaster last updated on 06/Feb/25 $$\mathrm{Let}\:{u}={x}+\mathrm{1}\Rightarrow{du}={dx},{x}={u}−\mathrm{1} \\ $$$$\int\frac{\left({u}−\mathrm{1}\right){e}^{{u}−\mathrm{1}} }{{u}^{\mathrm{2}}…