Question

If ${\int }_0^{\infty }\frac{x^3}{(a^2+x^2{)}^5}dx=\frac{1}{k{a}^6}$, then find the value of $\frac{k}{8}$

Answer 1

Let $x=atant$

$\therefore dx=ase{c}^{2}tdt$

$\therefore$ we have

${\int }_0^{\infty }\frac{x^3}{(a^2+x^2{)}^5}dx={\int }_0^{\frac{\pi }{2}}\frac{(atant{)}^3}{(a^2+(atant{)}^2{)}^5}ase{c}^{2}tdt$

$={\int }_0^{\frac{\pi }{2}}\frac{a^{4}ta{n}^{3}tse{c}^{2}t}{a^{10}(1+ta{n}^{2}t{)}^5}dt$

$={\int }_0^{\frac{\pi }{2}}\frac{ta{n}^{3}tse{c}^{2}t}{a^{6}se{c}^{10}t}dt$

$=\frac{1}{a^6}{\int }_0^{\frac{\pi }{2}}\frac{si{n}^{3}tco{s}^{10}t}{co{s}^{5}t}dt$

$=\frac{1}{a^6}{\int }_0^{\frac{\pi }{2}}si{n}^{3}t(1-si{n}^{2}t{)}^{2}costdt$

Now, let $sint=\theta$

$\therefore costdt=d\theta$

$\frac{1}{a^6}{\int }_0^{\frac{\pi }{2}}si{n}^{3}t(1-si{n}^{2}t{)}^{2}costdt=\frac{1}{a^6}{\int }_0^1{\theta }^3(1-{\theta }^2{)}^{2}d\theta$

$=\frac{1}{a^6}{\int }_0^1({\theta }^3-2{\theta }^5+{\theta }^7)d\theta$

$=\frac{1}{a^6}(\frac{{\theta }^4}{4}-\frac{2{\theta }^6}{6}+\frac{{\theta }^8}{8}){|}_0^1$

$=\frac{1}{a^6}(\frac{1}{4}-\frac{1}{3}+\frac{1}{8})$

$=\frac{1}{24a^6}$

$⟹k=24⟹\frac{k}{8}=3$