Question

Evaluate: ${\int }_0^{\infty }\frac{dx}{(1+x^2{)}^4}$

• A. $\frac{\pi }{32}$
• B. $\frac{3\pi }{32}$
• C. $\frac{5\pi }{32}$
• D. $\frac{7\pi }{32}$

Answer 1

Answer: (C)

$\frac{5\pi }{32}$

Formulae used;

${\int }_{\varphi =0}^{\varphi =n\pi }{\sin }^2\varphi d\varphi =\frac{n\pi }{2}$ (remember this trick!)...(same for cosine function also).

$\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$......(i)

$1+{\tan }^2\theta ={\sec }^2\theta$.....(ii)

${\sin }^6\theta +{\cos }^6\theta =1-3{\sin }^2\theta {\cos }^2\theta$.....(iii)

Proof for the last one,

${({\sin }^2\theta +{\cos }^2\theta )}^3={\sin }^6\theta +{\cos }^6\theta +3{\sin }^2\theta {\cos }^2\theta ({\sin }^2\theta +{\cos }^2\theta )\therefore 1={\sin }^6\theta +{\cos }^6\theta +3{\sin }^2\theta {\cos }^2\theta \therefore {\sin }^6\theta +{\cos }^6\theta =1-3{\sin }^2\theta {\cos }^2\theta$

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$I={\int }_0^{\infty }\frac{dx}{{(1+x^2)}^4}$

Let $x=\tan \left(\theta \right)$ and limits of the integral be $\theta =0$ to $\theta =\frac{\pi }{2}$

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\frac{d\tan \theta }{{(1+{\tan }^2\theta )}^4}={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\theta d\theta }{{\left({\sec }^2\theta \right)}^4}={\int }_0^{\frac{\pi }{2}}{\cos }^6\theta d\theta I={\int }_0^{\frac{\pi }{2}}{\cos }^6(\frac{\pi }{2}-\theta )d\theta ={\int }_0^{\frac{\pi }{2}}{\sin }^6\theta d\theta 2I={\int }_0^{\frac{\pi }{2}}{\sin }^6\theta d\theta +{\int }_0^{\frac{\pi }{2}}{\cos }^6\theta d\theta ={\int }_0^{\frac{\pi }{2}}{\sin }^6\theta +{\cos }^6\theta d\theta 2I={\int }_0^{\frac{\pi }{2}}1-3{\sin }^2\theta {\cos }^2\theta d\theta ={\int }_0^{\frac{\pi }{2}}d\theta -3{\int }_0^{\frac{\pi }{2}}{\sin }^2\theta {\cos }^2\theta d\theta 2I=\frac{\pi }{2}-3\left({\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}2\theta }{4}d\theta \right)let\theta =\frac{\varphi }{2}2I=\frac{\pi }{2}-\frac{3}{8}\left({\int }_{\varphi =0}^{\varphi =\pi }{\sin }^2\varphi d\varphi \right)$

$2I=\frac{\pi }{2}-\frac{3}{8}\left(\frac{\pi }{2}\right)=\frac{5\pi }{16}\therefore I=\frac{5\pi }{32}$