Question

The value of the integral $I=\displaystyle \int _{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d x}{1 + x^{2} + x^{3} + x^{5}}$ 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 $I=\displaystyle \int _{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d x}{\left(1 + x^{2}\right) \left(1 + x^{3}\right)}$
Let, $tan^{- 1} x=\theta $
$\Rightarrow dx=sec^{2} \theta d\theta $
$\therefore I=\displaystyle \int _{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{d \theta }{1 + tan^{3} \theta }$
$=\displaystyle \int _{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{cos^{3} \theta }{sin^{3} ⁡ \theta + cos^{3} ⁡ \theta }d\theta $
Applying $\left(a + b - x\right)$ property and adding, we get,
$2I=\displaystyle \int _{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{cos^{3} \theta + sin^{3} ⁡ \theta }{sin^{3} ⁡ \theta + cos^{3} ⁡ \theta }d\theta $
$2I=\left[\theta \right]_{\frac{\pi }{6}}^{\frac{\pi }{3}}$
$\Rightarrow 2I=\frac{\pi }{6}$
$\Rightarrow I=\frac{\pi }{12}$