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Category: Integration

Question-100450

Question Number 100450 by 175 last updated on 26/Jun/20 Answered by maths mind last updated on 26/Jun/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right)^{{a}} }{\:\sqrt{{x}}.\sqrt{\mathrm{1}−{x}}.\sqrt{\mathrm{1}+{x}}}{dx}={f}\left({a}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}}…

pi-2-pi-2-cos-2n-1-x-cos-2n-1-x-dx-2cos-2n-1-2-x-2n-1-pi-2-pi-2-0-What-is-the-mistake-in-above-pi-2-pi-2-cos-2n-1-x-cos-2n-1-x-dx-2-0-pi-2-co

Question Number 34901 by rishabh last updated on 12/May/18 $$\int_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}−\mathrm{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}}{dx} \\ $$$$=\left[−\frac{\mathrm{2cos}^{\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}} {x}}{\mathrm{2}{n}+\mathrm{1}}\right]_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} =\mathrm{0}? \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{above}? \\ $$$$\int_{−\pi/\mathrm{2}} ^{+\pi/\mathrm{2}} \sqrt{\mathrm{cos}^{\mathrm{2}{n}−\mathrm{1}}…

C-0-pi-dx-2-cos-2x-

Question Number 165955 by cortano1 last updated on 10/Feb/22 $$\:\:\:\:\mathrm{C}\:=\:\int_{\mathrm{0}} ^{\:\pi} \frac{\mathrm{dx}}{\mathrm{2}+\mathrm{cos}\:\mathrm{2x}}\:=? \\ $$ Answered by MJS_new last updated on 10/Feb/22 $$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{\mathrm{2}+\mathrm{cos}\:\mathrm{2}{x}}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}}…