Menu Close

Category: Integration

sin-1-x-2-dx-

Question Number 122877 by bemath last updated on 20/Nov/20 $$\:\:\int\:\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} \:{dx}\:? \\ $$ Commented by liberty last updated on 20/Nov/20 $$\:{let}\:{u}\:=\:\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow{x}\:=\:\mathrm{sin}\:{u}\: \\ $$$$\Rightarrow\:{dx}\:=\:\mathrm{cos}\:{u}\:{du}\:…

Question-122867

Question Number 122867 by bemath last updated on 20/Nov/20 Answered by som(math1967) last updated on 20/Nov/20 $$\mathrm{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\sqrt{\mathrm{cosx}}\mathrm{dx}}{\:\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}}} \\ $$$$\mathrm{again}\:\mathrm{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\sqrt{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}\mathrm{dx}}{\:\sqrt{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}+\sqrt{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}} \\ $$$$=\underset{\mathrm{0}}…

we-want-to-find-the-vslue-of-I-0-1-ln-1-x-1-x-2-dx-let-A-W-x-1-x-2-1-xy-dxdy-with-W-0-1-2-calculate-A-by-two-method-and-conclude-the-value-of-I-

Question Number 57324 by turbo msup by abdo last updated on 02/Apr/19 $${we}\:{want}\:{to}\:{find}\:{the}\:{vslue}\:{of} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:{let} \\ $$$${A}=\int\int_{{W}} \frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}{dxdy} \\ $$$${with}\:{W}=\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} \\…

calculate-D-x-y-3-x-2-y-2-dxdy-with-D-x-y-R-2-x-2-y-2-2-and-x-0-y-0-

Question Number 57323 by turbo msup by abdo last updated on 02/Apr/19 $${calculate}\:\int\int_{{D}} \:\:\frac{{x}+{y}}{\mathrm{3}+\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}\right. \\ $$$$\left.{and}\:{x}\geqslant\mathrm{0}\:,{y}\geqslant\mathrm{0}\right\} \\ $$…