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Question Number 87325 by M±th+et£s last updated on 04/Apr/20

1) ∫e^(√x)  dx  2)∫((√(sin(x)))/((√(sin(x)))+(√(cos(x)))))dx

$$\left.\mathrm{1}\right)\:\int{e}^{\sqrt{{x}}} \:{dx} \\ $$$$\left.\mathrm{2}\right)\int\frac{\sqrt{{sin}\left({x}\right)}}{\sqrt{{sin}\left({x}\right)}+\sqrt{{cos}\left({x}\right)}}{dx} \\ $$

Commented by Serlea last updated on 04/Apr/20

1) ∫e^(√x) dx  let (√x)=Y  x=y^2   dx=y^2 dy   ∫e^(√x) dx=∫e^y (2ydy)  =∫2ye^y dy  U=2y U′=2  V′=e^y       V=e^y     ∫2ye^y dy=2Ye^y −∫2e^y                      =2Ye^y −2e^y                      =2e^y (y−1)                    =2e^(√x) ((√x)−1)

$$\left.\mathrm{1}\right)\:\int\mathrm{e}^{\sqrt{\mathrm{x}}} \mathrm{dx} \\ $$$$\mathrm{let}\:\sqrt{\mathrm{x}}=\mathrm{Y} \\ $$$$\mathrm{x}=\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{dx}=\mathrm{y}^{\mathrm{2}} \mathrm{dy} \\ $$$$\:\int\mathrm{e}^{\sqrt{\mathrm{x}}} \mathrm{dx}=\int\mathrm{e}^{\mathrm{y}} \left(\mathrm{2ydy}\right) \\ $$$$=\int\mathrm{2ye}^{\mathrm{y}} \mathrm{dy} \\ $$$$\mathrm{U}=\mathrm{2y}\:\mathrm{U}'=\mathrm{2} \\ $$$$\mathrm{V}'=\mathrm{e}^{\mathrm{y}} \:\:\:\:\:\:\mathrm{V}=\mathrm{e}^{\mathrm{y}} \\ $$$$ \\ $$$$\int\mathrm{2ye}^{\mathrm{y}} \mathrm{dy}=\mathrm{2Ye}^{\mathrm{y}} −\int\mathrm{2e}^{\mathrm{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2Ye}^{\mathrm{y}} −\mathrm{2e}^{\mathrm{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2e}^{\mathrm{y}} \left(\mathrm{y}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2e}^{\sqrt{\mathrm{x}}} \left(\sqrt{\mathrm{x}}−\mathrm{1}\right) \\ $$$$ \\ $$

Commented by peter frank last updated on 04/Apr/20

qn 2 limit ?

$${qn}\:\mathrm{2}\:{limit}\:? \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 04/Apr/20

no sir

$${no}\:{sir} \\ $$

Answered by MJS last updated on 04/Apr/20

2)  ∫((√(sin x))/((√(sin x))+(√(cos x))))dx=       [t=(√(tan x)) → dx=2cos^2  x (√(tan x))dt]  =2∫(t^2 /((t^4 +1)(t+1)))dt=  =−((1−(√2))/2)∫((t+1)/(t^2 −(√2)t+1))dt−((1+(√2))/2)∫((t+1)/(t^2 +(√2)t+1))dt+∫(dt/(t+1))  now it should be easy

$$\left.\mathrm{2}\right) \\ $$$$\int\frac{\sqrt{\mathrm{sin}\:{x}}}{\sqrt{\mathrm{sin}\:{x}}+\sqrt{\mathrm{cos}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{4}} +\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{dt}= \\ $$$$=−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}+\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\ $$

Commented by peter frank last updated on 04/Apr/20

thank you

$${thank}\:{you} \\ $$

Commented by M±th+et£s last updated on 04/Apr/20

thank you

$${thank}\:{you} \\ $$

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