Question

The value of $k>0$ such that ${\int }_0^k\left[{\tan }^{-1}\sqrt{x}\right]dx$ $={\int }_0^k\left[co{t}^1\sqrt{x}\right]dx$ where $[.]$ denote the greatest integer function, is

• A. $\frac{3-4cos4}{1-\cos 4}$
• B. $2\left(\frac{3+4cos4}{1-\cos 4}\right)$
• C. $\frac{1-\cos 4}{3+\cos 4}$
• D. $\frac{2(1-\cos 4)}{3+\cos 4}$

Answer 1

Answer: (B)

$2\left(\frac{3+4cos4}{1-\cos 4}\right)$

The given interval will be zero if $k<{\tan }^{2}1$

$\therefore {\int }_0^k{\tan }^{-1}\sqrt{x}dx={\int }_0^k{\cot }^{-1}\sqrt{x}dx$

$\Rightarrow {\int }_{{\tan }^{2}1}^{k}1dx=-{\int }_{{\cot }^{2}1}^{0}dx={\int }_0^{{\cot }^{2}1}dx$

$\Rightarrow k-{\tan }^{2}1={\cot }^{2}1$

$\Rightarrow k={\cot }^{2}1+{\tan }^{2}1$

$=\frac{{({\sin }^{2}1+{\cos }^{2}1)}^2-2{\sin }^{2}1{\cos }^{2}1}{{\sin }^{2}1{\cos }^{2}1}$

$=\frac{4[1-\frac{1}{2}{\sin }^{2}2]\times 2}{2{\sin }^{2}2}$

$=2\left(\frac{3+\cos 4}{1-\cos 4}\right)$