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Question Number 35832 by abdo mathsup 649 cc last updated on 24/May/18

$$findthevalueoff\left(\lambda \right)={\int }_{-a}^a\frac{dx}{{\left(\lambda {+}_{}{x}^2\right)}^{\frac{3}{2}}}$$$$\lambda \in R.$$

Commented by abdo.msup.com last updated on 25/May/18

$$case1if\lambda >0letusethechangement$$$$x=\sqrt{\lambda }tan\left(t\right)$$$$f\left(\lambda \right)={\int }_{-arctan\left(\frac{a}{\lambda }\right)}^{arctan\left(\frac{a}{\lambda }\right)}\frac{\sqrt{\lambda }}{{\lambda }^{\frac{3}{2}}{(1+{tan}^2\left(t\right))}^{\frac{3}{2}}}(1+{tan}^{2}t)dt$$$$=\frac{1}{\lambda }{\int }_{-arctan\left(\frac{a}{\sqrt{\lambda }}\right)}^{arctan\left(\frac{a}{\sqrt{\lambda }}\right)}\frac{1}{{(1+{tan}^{2}t)}^{\frac{1}{2}}}dt$$$$=\frac{2}{\lambda }{\int }_0^{arctan\left(\frac{a}{\sqrt{\lambda }}\right)}cos\left(t\right)dt$$$$=\frac{2}{\lambda }{\left[sin\left(t\right)\right]}_0^{arctan\left(\frac{a}{\sqrt{\lambda }}\right)}$$$$=\frac{2}{\lambda }sin\left(arctan\left(\frac{a}{\sqrt{\lambda }}\right)\right)$$$$butwehavesin\left(arctanx\right)=\frac{x}{\sqrt{1+x^2}}$$$$sosin\left(arctan\left(\frac{a}{\sqrt{\lambda }}\right)\right)=\frac{a}{\sqrt{\lambda }\sqrt{1+\frac{a^2}{\lambda }}}$$$$=\frac{a}{\sqrt{\lambda +a^2}}\Rightarrow f\left(\lambda \right)=\frac{2a}{\lambda \sqrt{\lambda +a^2}}$$

Commented by prof Abdo imad last updated on 25/May/18

$$case2if\lambda <0\Rightarrow -\lambda >0letput\alpha =-\lambda$$$$f\left(\lambda \right)={\int }_{-a}^a\frac{dx}{{(x^2-\alpha )}^{\frac{3}{2}}}.changementx=\sqrt{\alpha }ch\left(t\right)$$$$give$$$$f\left(\lambda \right)={\int }_{argch\left(\frac{-a}{\sqrt{\alpha }}\right)}^{argch\left(\frac{a}{\sqrt{\alpha }}\right)}\frac{\sqrt{\alpha }sh\left(t\right)}{{\alpha }^{\frac{3}{2}}{({ch}^2\left(t\right)-1)}^{\frac{3}{2}}}dt$$$$=\frac{1}{\alpha }{\int }_{argch\left(\frac{-a}{\sqrt{\alpha }}\right)}^{argch\left(\frac{a}{\sqrt{\alpha }}\right)}\frac{sh\left(t\right)}{{sh}^3\left(t\right)}dt$$$$=\frac{2}{\alpha }{\int }_{argch\left(\frac{-a}{\sqrt{\alpha }}\right)}^{argch\left(\frac{a}{\sqrt{\alpha }}\right)}\frac{dt}{ch\left(2t\right)-1}$$$$=\frac{2}{\alpha }{\int }_{ln(-\frac{a}{\sqrt{\kappa }}+\sqrt{\frac{a^2}{\alpha }-1})}^{ln(\frac{a}{\sqrt{\alpha }}+\sqrt{\frac{a^2}{\alpha }-1}}\frac{dt}{\frac{e^{2t}+e^{-2t}}{2}-1}$$$$=\frac{4}{\alpha }{\int }_{ln(-\frac{a}{\sqrt{\alpha }}+\sqrt{\frac{a^2}{\alpha }-1})}^{ln(\frac{a}{\sqrt{\alpha }}+\sqrt{\frac{a^2}{\alpha }-1})}\frac{dt}{e^{2t}+e^{-2t}-2}...$$$$becontinued...$$$$$$

Commented by prof Abdo imad last updated on 26/May/18

$$letfindK=\int \frac{dt}{e^{2t}+e^{-2t}-2}chamgement$$$$e^{2t}=ugiveK=\int \frac{1}{u+u^{-1}-2}\frac{du}{2u}$$$$=\frac{1}{2}\int \frac{du}{u^2+1-2u}=\frac{1}{2}\int \frac{du}{{(u-1)}^2}$$$$=-\frac{1}{2(u-1)}=\frac{-1}{2(e^{2t}-1)}+cso$$$$f\left(\lambda \right)=-\frac{4}{2\alpha }{\left[\frac{1}{e^{2t}-1}\right]}_{ln(\frac{-a}{\sqrt{\alpha }}+\sqrt{\frac{a^2}{\alpha }-1}}^{ln(\frac{a}{\sqrt{\alpha }}+\sqrt{\frac{a^2}{\alpha }-1}}$$$$=\frac{-2}{\alpha }\{\frac{1}{(\frac{a}{\sqrt{\alpha }}+\sqrt{\frac{a^2}{\alpha }-1})}-\frac{1}{(-\frac{a}{\sqrt{\alpha }}+\sqrt{\frac{a^2}{\alpha }-1})}$$$$=\frac{2}{\lambda }\{\frac{1}{(\frac{a}{\sqrt{-\lambda }}+\sqrt{-\frac{a^2}{\lambda }-1})}-\frac{1}{(\frac{-a}{\sqrt{-\lambda }}+\sqrt{-\frac{a^2}{\lambda }-1})}$$$$with\lambda <0.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18

$$x=\sqrt{\lambda }tan\theta dx=\sqrt{\lambda }{sec}^2\theta d\theta$$$$I=\int \frac{\sqrt{\lambda }{sec}^2\theta d\theta }{{\left\{\lambda (1+{tan}^2\theta )\right\}}^{\frac{3}{2}}}$$$$\int \frac{1}{\lambda }\times \frac{{sec}^2\theta }{{sec}^3\theta }\times \frac{d\theta }{}$$$$=\frac{1}{\lambda }\int cos\theta d\theta$$$$=\frac{1}{\lambda }sin\theta nowputtingthelimit$$$$=\frac{1}{\lambda }\times \mid \frac{x}{{(\lambda +x^2)}^{\frac{1}{2}}}{\mid }_{-a}^a$$$$=\frac{1}{\lambda }\times \frac{2a}{{(\lambda +a^2)}^{\frac{1}{2}}}$$

Commented by abdo.msup.com last updated on 25/May/18

$$youransweriscorrectsirTanmay$$$$butyoumuststudythecase\lambda <0....$$

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