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Question Number 42793 by maxmathsup by imad last updated on 02/Sep/18

$$findf\left(a\right)={\int }_0^1\frac{dt}{{(a^2+t^2)}^3}witha>0$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

$$t=atan\alpha dt={asec}^2\alpha d\alpha$$ $${\int }_0^{{tan}^{-1}\left(\frac{1}{a}\right)}\frac{{asec}^2\alpha d\alpha }{{\left\{a^2(1+{tan}^2\alpha )\right\}}^3}$$ $${\int }_0^{{tan}^{-1}\left(\frac{1}{a}\right)}\frac{{asec}^2\alpha d\alpha }{a^6{sec}^6\alpha }$$ $$\frac{1}{a^5}{\int }_0^{{tan}^{-1}\left(\frac{1}{a}\right)}{\left(\frac{1+cos2\alpha }{2}\right)}^{2}d\alpha$$ $$\frac{1}{4a^5}{\int }_0^{{tan}^{-1}\left(\frac{1}{a}\right)}1+2cos2\alpha +\frac{1+cos4\alpha }{2}d\alpha$$ $$\frac{1}{8a^5}{\int }_0^{{tan}^{-1}\left(\frac{1}{a}\right)}3+4cos2\alpha +cos4\alpha d\alpha$$ $$\frac{1}{8a^5}\mid (3\alpha +\frac{4sin2\alpha }{2}+\frac{sin4\alpha }{4}){\mid }_0^{{tan}^{-1}\left(\frac{1}{a}\right)}$$ $$nowtan\alpha =\frac{1}{a}$$ $$sin2\alpha =\frac{2tan\alpha }{1+{tan}^2\alpha }=\frac{\frac{2}{a}}{1+\frac{1}{a^2}}=\frac{2a}{1+a^2}$$ $$sin4\alpha =2sin2\alpha .cos2\alpha$$ $$=2\left(\frac{2t}{1+t^2}\times \frac{1-t^2}{1+t^2}\right)$$ $$=\frac{4\left(\frac{1}{a}\right)(1-\frac{1}{a^2})}{{(1+\frac{1}{a^2})}^2}=\frac{\frac{4(a^2-1)}{a^3}}{\frac{{(a^2+1)}^2}{a^4}}=\frac{4a(a^2-1)}{{(a^2+1)}^2}$$ $$nowputthevalue...$$ $$contdsotheansis$$ $$=\frac{1}{8a^5}\mid 3{tan}^{-1}\left(\frac{1}{a}\right)+2.\frac{2a}{1+a^2}+\frac{1}{4}.\frac{4a(a^2-1)}{{(a^2+1)}^2}\mid$$ $$=\frac{1}{8a^5}\{3{tan}^{-1}\left(\frac{1}{a}\right)+\frac{4a}{1+a^2}+\frac{a^3-a}{{(a^2+1)}^2}\}$$

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