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Category: Integration

BeMath-x-x-2a-x-dx-

Question Number 107965 by bemath last updated on 13/Aug/20 $$\:\:\:\:\frac{\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{} \\ $$$$\int\:{x}\:\sqrt{\frac{{x}}{\mathrm{2}{a}−{x}}}\:{dx}\:?\: \\ $$ Answered by bobhans last updated on 13/Aug/20 $$\:\:\frac{\ddots\mathscr{BOBHANS}\iddots}{\Pi} \\ $$$$\:\mathrm{I}=\:\int\:\mathrm{x}\:\sqrt{\frac{\mathrm{x}}{\mathrm{2a}−\mathrm{x}}}\:\mathrm{dx}\:\:\left[\:\mathrm{set}\:\mathrm{x}\:=\:\mathrm{a}−\mathrm{acos}\:\mathrm{2t}\:\right] \\…

0-cos-ax-sin-bx-x-2-dx-0-1-2sin-2-ax-2-1-2sin-2-bx-2-x-2-dx-2-0-sin-bx-2-x-2-dx-1-2-0-

Question Number 173486 by mnjuly1970 last updated on 12/Jul/22 $$ \\ $$$$\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\:\left({ax}\right)−\:{sin}\left({bx}\right)}{{x}^{\:\mathrm{2}} }{dx}\:\:=? \\ $$$$\:\:\: \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{ax}}{\mathrm{2}}\right)−\left(\mathrm{1}−\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{bx}}{\mathrm{2}}\right)\right)}{{x}\:^{\mathrm{2}} }{dx} \\…

1-1-tanx-dx-

Question Number 42407 by Tawa1 last updated on 25/Aug/18 $$\int\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{tanx}}\:\mathrm{dx} \\ $$ Commented by maxmathsup by imad last updated on 25/Aug/18 $${let}\:{I}\:\:=\:\int\:\:\:\:\:\frac{{dx}}{\mathrm{1}+{tanx}}\:\:{changement}\:{tanx}\:={t}\:{give} \\ $$$${I}\:\:=\:\int\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\:\:{let}\:{decompose}\:{F}\left({t}\right)\:=\:\:\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}}…

calculate-D-xy-1-x-2-y-2-dxdy-with-D-x-y-R-2-0-x-1-0-y-1-x-2-y-2-1-

Question Number 42395 by abdo.msup.com last updated on 24/Aug/18 $${calculate}\:\int\int_{{D}} \:\:\:\:\:\:\frac{{xy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)}{dxdy}\:{with} \\ $$$${D}\:=\left\{\left({x},{y}\right)\in\:{R}^{\mathrm{2}} \:\:/\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:,\mathrm{0}\leqslant{y}\leqslant\mathrm{1},\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{1}\right\} \\ $$ Commented by maxmathsup by imad…