Question

The value of the integral $I={\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{dx}{1+x^2+x^3+x^5}$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{12}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (C)

Given integral is $I={\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{dx}{\left(1+x^2\right)\left(1+x^3\right)}$
Let, $ta{n}^{-1}x=\theta$
$\Rightarrow dx=se{c}^2\theta d\theta$
$\therefore I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{d\theta }{1+ta{n}^3\theta }$
$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{co{s}^3\theta }{si{n}^3\theta +co{s}^3\theta }d\theta$
Applying $\left(a+b-x\right)$ property and adding, we get,
$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{co{s}^3\theta +si{n}^3\theta }{si{n}^3\theta +co{s}^3\theta }d\theta$
$2I={\left[\theta \right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}$
$\Rightarrow 2I=\frac{\pi }{6}$
$\Rightarrow I=\frac{\pi }{12}$