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Category: Integration

plzz-help-ne-differentiate-between-sin-2x-1-2-cox-2x-c-is-not-change-to-2sin-x-cos-x-but-b-a-sin-2x-is-change-to-b-a-2sin-x-cos-x-

Question Number 32627 by Cheyboy last updated on 29/Mar/18 $${plzz}\:{help}\:{ne}\:{differentiate}\: \\ $$$${between} \\ $$$$\int{sin}\left(\mathrm{2}{x}\right)=\:−\frac{\mathrm{1}}{\mathrm{2}}{cox}\left(\mathrm{2}{x}\right)+{c}\: \\ $$$${is}\:{not}\:{change}\:{to}\:\int\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right) \\ $$$${but}\:\underset{{b}} {\overset{{a}} {\int}}{sin}\left(\mathrm{2}{x}\right)=\:{is}\:{change}\:{to} \\ $$$$\underset{{b}} {\overset{{a}} {\int}}\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right) \\…

e-x-5-8x-2-dx-pi-4-2-e-x-5-erfi-2-2-x-5-pi-4-128-2-super-erf-hyper-2-2-x-c-where-super-erf-hyper-t-is-super-function-in-D-2-and-D-n-

Question Number 98151 by  M±th+et+s last updated on 11/Jun/20 $$\int{e}^{{x}^{\mathrm{5}} +\mathrm{8}{x}^{\mathrm{2}} } {dx} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}{e}^{{x}^{\mathrm{5}} } {erfi}\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)−\frac{\mathrm{5}\sqrt{\pi}}{\mathrm{4}\left(\mathrm{128}\right)\sqrt{\mathrm{2}}}\left({super}−{erf}_{\left({hyper}\right)} \left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)\right)+{c} \\ $$$$ \\ $$$${where}\left[{super}−{erf}_{\left({hyper}\right)} \left({t}\right)\right]\:{is}\:{super}−{function} \\ $$$${in}\:{D}_{\mathrm{2}}…

Question-163573

Question Number 163573 by Zaynal last updated on 08/Jan/22 Commented by MJS_new last updated on 08/Jan/22 $$=\frac{\mathrm{sec}^{−\mathrm{1}} }{\mathrm{3cosec}^{−\mathrm{1}} }\int\frac{{dx}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{or}\:\mathrm{rewrite}\:\mathrm{it}\:\mathrm{properly} \\ $$ Terms…

What-is-the-area-of-the-region-bounded-by-x-2-y-2-9-x-y-3-and-y-x-

Question Number 98020 by bobhans last updated on 11/Jun/20 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded} \\ $$$$\mathrm{by}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\leqslant\:\mathrm{9}\:;\:\mathrm{x}+\mathrm{y}\:\leqslant\:\mathrm{3}\:\mathrm{and}\:\mathrm{y}\:\leqslant\:\mathrm{x}\: \\ $$ Answered by bemath last updated on 11/Jun/20 Commented by…