Menu Close

Category: Integration

dx-sin-4-x-

Question Number 158053 by Ar Brandon last updated on 30/Oct/21 $$\int\frac{{dx}}{\mathrm{sin}^{\mathrm{4}} {x}} \\ $$ Answered by ajfour last updated on 30/Oct/21 $$\int\frac{{dx}}{\left[\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right]^{\mathrm{2}} }=\int\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt}}{{t}^{\mathrm{4}}…

Question-26952

Question Number 26952 by Femmy last updated on 31/Dec/17 Answered by prakash jain last updated on 31/Dec/17 $$\int\frac{\mathrm{sin}^{\mathrm{5}} {x}}{\mathrm{cos}^{\mathrm{3}} {x}}{dx}=\int\frac{\mathrm{sin}^{\mathrm{4}} {x}}{\mathrm{cos}^{\mathrm{3}} {x}}\mathrm{sin}\:{xdx} \\ $$$$=\int\frac{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}}…

0-1-6x-2-x-1-6x-2-5x-1-dx-

Question Number 26950 by Joel578 last updated on 31/Dec/17 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{6}{x}^{\mathrm{2}} \:−\:{x}\:−\:\mathrm{1}}{\mathrm{6}{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{1}}\:{dx} \\ $$ Commented by prakash jain last updated on 31/Dec/17 $$\mathrm{Integral}\:\mathrm{does}\:\mathrm{not}\:\mathrm{converge}.…

5-7-x-2-f-x-dx-if-f-7-2-f-5-1-f-7-2-f-5-1-f-7-3-f-5-4-

Question Number 92484 by jagoll last updated on 07/May/20 $$\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{f}\:'''\left(\mathrm{x}\right)\:\mathrm{dx}\:=? \\ $$$$\mathrm{if}\:\mathrm{f}''\left(\mathrm{7}\right)\:=\:\mathrm{2}\:,\:\mathrm{f}''\left(\mathrm{5}\right)\:=\mathrm{1} \\ $$$$\mathrm{f}'\left(\mathrm{7}\right)\:=\:−\mathrm{2}\:,\:\mathrm{f}'\left(\mathrm{5}\right)\:=\:−\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{7}\right)=\:−\mathrm{3}\:,\:\mathrm{f}\left(\mathrm{5}\right)\:=\:−\mathrm{4}\: \\ $$$$ \\ $$ Answered by…

dx-1-x-1-4-x-

Question Number 157977 by tounghoungko last updated on 30/Oct/21 $$\int\:\frac{{dx}}{\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{{x}}\:\right)\sqrt{{x}}}\:=? \\ $$ Answered by puissant last updated on 30/Oct/21 $$\Omega=\int\frac{{dx}}{\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{{x}}\right)\sqrt{{x}}} \\ $$$${u}=\sqrt{{x}\:}\:\rightarrow\:{du}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}\:\rightarrow\:{dx}=\:\mathrm{2}\sqrt{{x}}{du} \\ $$$$\Rightarrow\:\Omega\:=\:\int\frac{\mathrm{2}\sqrt{{x}}{du}}{\left(\mathrm{1}+\sqrt{{u}}\right)\sqrt{{x}}}\:=\:\mathrm{2}\int\frac{{du}}{\left(\mathrm{1}+\sqrt{{u}}\right)} \\…