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Definite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals1 Mark
Question
Given that $\int\limits^{\infty}_0\frac{\text{x}^2}{(\text{x}^2+\text{a}^2)(\text{x}^2+\text{b}^2)(\text{x}^2+\text{c}^2)}\text{ dx}=\frac{\pi}{2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})},$ the value of $\int\limits^\infty_0\frac{1}{(\text{x}^2+4)(\text{x}^2+9)},$ is:

  1. $\frac{\pi}{60}$

  2. $\frac{\pi}{20}$

  3. $\frac{\pi}{40}$

  4. $\frac{\pi}{80}$

✓

Answer

  1. $\frac{\pi}{60}$

Solution:

$\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)\big(\text{x}^2+9\big)}\text{dx}$

$=\frac{1}{5}\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)}-\frac{1}{\big(\text{x}^2+9\big)}\text{dx}$

$=\frac{1}{5}\bigg[\frac{1}{2\tan^{-1}{}}\frac{\text{x}}{2}-\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}\bigg]^\infty_0$

$=\frac{1}{5}\bigg[\frac{1}{2}\times\frac{\pi}{2}-\frac{1}{3}\times\frac{\pi}{2}\bigg]$

$=\frac{1}{5}\times\frac{\pi}{12}$

$=\frac{\pi}{60}$

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