Question

$\int _{\pi/6}^{\pi/4} \frac {1}{\sqrt{\cos\,x \,\sin^3\,x}}dx$ is equal to

• A. 1
• B. 1/3
• C. -2
• D. none of these

Answer 1

Answer: (D)

Let $I=\int\limits_{\pi /6}^{\pi /4} \frac{dx}{\sqrt{cos\,x \, sin^{3}\,x}}$
$=\int\limits_{\pi/ 6}^{\pi /4} \frac{sec^{2}\,x}{\sqrt{tan^{3}\,x}} dx$
$=\left|\frac{\left(tan\,x\right)^{-1 /2}}{-1 /2}\right|_{\pi/ 6}^{\pi /4}$
$=-2\left[\left(1\right)-\left(\frac{1}{\sqrt{3}}\right)^{-1 /2}\right]$
$=-2\left[1-\left(\sqrt{3}\right)^{1 /2}\right]$
$=2\left(3\right)^{1 /4}-2$