Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 126274 by bramlexs22 last updated on 19/Dec/20

Answered by liberty last updated on 19/Dec/20

 letting x=(√2) sec ℓ with → { ((ℓ=(π/2))),((ℓ=(π/4))) :}   ∫_(π/4) ^( π/2)  (((√2) sec ℓ tan ℓ)/( (√2) sec ℓ (√(2tan^2 ℓ)))) dℓ =  (1/( (√2)))∫ dℓ = (1/( (√2))) [ (π/2)−(π/4) ] = (π/(4(√2))) = ((π(√2))/8)

$$\:{letting}\:{x}=\sqrt{\mathrm{2}}\:\mathrm{sec}\:\ell\:{with}\:\rightarrow\begin{cases}{\ell=\frac{\pi}{\mathrm{2}}}\\{\ell=\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$$$\:\int_{\pi/\mathrm{4}} ^{\:\pi/\mathrm{2}} \:\frac{\sqrt{\mathrm{2}}\:\mathrm{sec}\:\ell\:\mathrm{tan}\:\ell}{\:\sqrt{\mathrm{2}}\:\mathrm{sec}\:\ell\:\sqrt{\mathrm{2tan}\:^{\mathrm{2}} \ell}}\:{d}\ell\:= \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:{d}\ell\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\:\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:\right]\:=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\: \\ $$

Answered by mathmax by abdo last updated on 19/Dec/20

I=∫_1 ^∞  (dx/(x(√(x^2 −2)))) we do the changement x=((√2)/(sint)) ⇒(dx/dt)=−(((√2)cost)/(sin^2 t)) ⇒  x^2 −2>0 ⇒x>(√2)or x<−(√2) ⇒ I is not defined...!

$$\mathrm{I}=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{sint}}\:\Rightarrow\frac{\mathrm{dx}}{\mathrm{dt}}=−\frac{\sqrt{\mathrm{2}}\mathrm{cost}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\:\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{2}>\mathrm{0}\:\Rightarrow\mathrm{x}>\sqrt{\mathrm{2}}\mathrm{or}\:\mathrm{x}<−\sqrt{\mathrm{2}}\:\Rightarrow\:\mathrm{I}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}...! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com