Question

Solve ${\int }_{\sqrt{2}}^1\frac{dx}{x^5\sqrt{x^2-1}}$

• A. $\frac{1}{32}(\pi +\frac{7\sqrt{3}}{2}-8)$
• B. $\frac{1}{32}(\pi -\frac{7\sqrt{3}}{2}-8)$
• C. $\frac{1}{32}(\pi +\frac{7\sqrt{3}}{2}+8)$
• D. None of these

Answer 1

The correct option is A $\frac{1}{32}(\pi +\frac{7\sqrt{3}}{2}-8)$

Let $I=\int \frac{1}{x^5\sqrt{x^2-1}}dx$
Put $x=\sec t\Rightarrow dx=\tan t\sec tdt$
$I=\int {\cos }^{4}tdt$
Using $\int {\cos }^{m}dx=\frac{\sin \left(x\right){\cos }^{m-1}x}{m}+\frac{m-1}{m}\int {\cos }^{-2+x}xdx$
$I=\frac{1}{4}\sin t{\cos }^{3}t+\frac{3}{4}\int {\cos }^{2}tdx$
$=\frac{1}{4}\sin t{\cos }^{3}t+\frac{3}{8}\int \left({\cos }^{2x}+\frac{1}{2}\right)dx$
$=\frac{3}{8}{\sec }^{-1}x+\frac{1}{4}\cos \left({\sec }^{-1}x\right)\sin \left({\sec }^{-1}x\right)+\frac{3}{8}\left(\sin \left({\sec }^{-1}x\right)\cos \left({\sec }^{-1}x\right)\right)$
$\therefore {\int }_{\sqrt{2}}^1\frac{dx}{x^5\sqrt{x^2-1}}=\frac{1}{32}(\pi +\frac{7\sqrt{3}}{2}-8)$