Question

${\int }_0^a\frac{x^{4}dx}{(a^2+x^2{)}^4}=$

• A. $\frac{1}{16a^3}(\frac{\pi }{4}-\frac{1}{3})$
• B. $\frac{1}{16a^3}(\frac{\pi }{4}+\frac{1}{3})$
• C. $\frac{1}{16}{a}^3(\frac{\pi }{4}-\frac{1}{3})$
• D. $\frac{1}{16}{a}^3(\frac{\pi }{4}+\frac{1}{3})$

Answer 1

The correct option is A $\frac{1}{16a^3}(\frac{\pi }{4}-\frac{1}{3})$

$Putx=atan\theta \Rightarrow dx=ase{c}^2\theta d\theta ,thenwehave$
$I={\int }_0^{\frac{\pi }{4}}\frac{a^{4}ta{n}^4\theta .ase{c}^2\theta d\theta }{a^{8}se{c}^8\theta }$
$\Rightarrow \frac{1}{a^3}{\int }_0^{\frac{\pi }{4}}si{n}^4\theta co{s}^2\theta d\theta =I=\frac{1}{a^3}\left[{\int }_0^{\frac{\pi }{4}}(si{n}^4\theta -si{n}^6\theta )\right]d\theta$
$=\frac{1}{a^3}{\int }_0^{\frac{\pi }{4}}[\frac{(1-cos2\theta {)}^2}{4}-\frac{(1-cos2\theta {)}^3}{8}]d\theta$
$=\frac{1}{8a^3}{\int }_0^{\frac{\pi }{4}}(1+cos2\theta )(1+co{s}^{2}2\theta -2cos2\theta )d\theta$
$=\frac{1}{8a^3}{\int }_0^{\frac{\pi }{4}}(1-cos2\theta -co{s}^{2}2\theta +co{s}^{3}2\theta )d\theta$
$=\frac{1}{32a^3}{\int }_0^{\frac{\pi }{4}}(2-cos2\theta -2cos4\theta +cos6\theta )d\theta$
$=\frac{1}{32a^3}{[2\theta -\frac{sin2\theta }{2}-\frac{sin4\theta }{2}+\frac{sin6\theta }{6}]}_0^{\frac{\pi }{4}}$
$=\frac{1}{16a^3}(\frac{\pi }{4}-\frac{1}{3})$