Question

The value of $\int^{^{^{\infty}}}_{_{_0}} \frac{dx}{\left(a^{2}+x^{2}\right)^{^7}}$ is equal to

• A. $\frac{231}{2047}\left(\frac{1}{a^{13}}\right)$
• B. $\frac{235}{2048}\left(\frac{1}{a^{13}}\right)$
• C. $\frac{232}{2047}\left(\frac{1}{a^{13}}\right)$
• D. $\frac{231}{2048}\left(\frac{1}{a^{13}}\right)$

Answer 1

Answer: (D)

Let I =$\int^{^{^{\infty}}}_{_{_0}} \frac{dx}{\left(a^{2}+x^{2}\right)^{^7}}$
Put $x = a tan\theta\Rightarrow dx = a sec2 \theta d\theta$
limit at x = $0 \Rightarrow \theta =0 \& x=\infty \Rightarrow \theta =\frac{\pi}{2}$
$\therefore I=\int^{^{\frac{\pi}{2}}}_{_0} \frac{a sec^{2} \theta}{a^{14}\left(1+tan^{2} \theta\right)^{7}}d\theta$
$=\frac{1}{_{a}13}\int^{^{\frac{\pi}{2}}}_{_0} \frac{1}{sec^{12} \theta }d\theta =\frac{1}{_{a}13}. \int^{^{\frac{\pi }{2}}}_{_0}cos^{12} \theta.d\theta$
But
$\int_{_0}^{^{\frac{\pi }{2}}} sin^{2m-1}\theta.d\theta=\frac{1}{2}B\left(m,n\right)$
$\& B\left(m,n\right)=\frac{mn}{m+n}\& \frac{1}{2}=\pi$
$\therefore I=\int^{^{\frac{\pi }{2}}}_{_0}sin^{0} \theta.cos^{12} \theta d\theta$
$\therefore m=\frac{1}{2}, n=\frac{13}{2}$
$\therefore I=\frac{1}{_{2a}13} \frac{\frac{1}{2}. \frac{13}{2}}{\frac{1}{2}+\frac{13}{2}}$
$=\frac{1}{_{2a}13}. \frac{\pi. \frac{11}{2}. \frac{9}{2}. \frac{7}{2}. \frac{5}{2}. \frac{3}{2}. \frac{1}{2}. \pi}{6.5.4.3.2.1}$
$=\left(\frac{231}{2048}. \frac{1}{a^{13}}\right)\pi$