Question

The value of $\int_{1}^{16}{\tan^{- 1}\sqrt{\sqrt{x} - 1}\mspace{6mu} dx}$ 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

Answer: (D)

$\frac{16}{3}\pi$– $2\sqrt{3}$

Answer 2

Integrating by parts, the given integral is equal to

x tan^–1 $\left. \ \sqrt{\sqrt{x} - 1} \right|_{1}^{16}$ $- \int_{1}^{16}{\frac{x}{\sqrt{x}}\frac{1}{4\sqrt{x}\sqrt{\sqrt{x} - 1}}}$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 + t²)

= $\frac{16}{3}\pi$ – $\left( \sqrt{3} + \sqrt{3} \right)$ = $\frac{16}{3}$p – 2$\sqrt{3}$