Question

The value of ${\int }_1^{16}ta{n}^{-1}\sqrt{\sqrt{x}-1}dx$ is

• A. $\frac{16\pi }{3}+2\sqrt{3}$
  
• B. $\frac{4}{3}\pi -2\sqrt{3}$
• C. $\frac{4}{3}\pi +2\sqrt{3}$
• D. $\frac{16}{3}\pi -2\sqrt{3}$

Answer 1

The correct option is D

$\frac{16}{3}\pi -2\sqrt{3}$

Integrating by parts, the given integral is equal to

$x{\begin{array}{c}ta{n}^{-1}\sqrt{\sqrt{x}-1}\end{array}|}_1^{16}-{\int }_1^{16}\frac{x}{\sqrt{x}}\frac{1}{4\sqrt{x}\sqrt{\sqrt{x}-1}}dx$

$=\frac{16}{3}\pi -\frac{1}{4}{\int }_1^{16}\frac{dx}{\sqrt{\sqrt{x}-1}}$

$=\frac{16}{3}\pi -\frac{1}{4}{\int }_0^{\sqrt{3}}\frac{4t(1+t^2)}{t}dt(\sqrt{x}=1+t^2)$

$\frac{16}{3}\pi -(\sqrt{3}+\sqrt{3})=\frac{16}{3}\pi -2\sqrt{3}$