The value of the constant $a>0$ such that $\int\limits_0^a\left[\tan ^{-1} \sqrt{x}\right] d x=\int\limits_0^a\left[\cot ^{-1} \sqrt{x}\right] d x$, where [.] denotes the greatest integer function, is
Answer & explanation
Correct answer: option 1
We have,
$\int\limits_0^a\left[\tan ^{-1} \sqrt{x}\right] d x=\int\limits_0^a\left[\cot ^{-1} \sqrt{x}\right] d x$
$\Rightarrow \int\limits_0^{\tan ^2 1}\left[\tan ^{-1} \sqrt{x}\right] d x+\int\limits_{\tan ^2 1}^a\left[\tan ^{-1} \sqrt{x}\right] d x$
$=\int\limits_0^{\cot ^2 1}\left[\cot ^{-1} \sqrt{x}\right] d x+\int\limits_{\cot ^2 1}^a\left[\cot ^{-1} \sqrt{x}\right] d x$
$\Rightarrow \int\limits_0^{\tan ^2 1} 0 d x+\int\limits_{\tan ^2 1}^a 1 d x=\int\limits_0^{\cot ^2 1} 1 . d x+\int\limits_{\cot ^2 1}^a 0 d x$
$\Rightarrow a-\tan ^2 1=\cot ^2 1$
$\Rightarrow a=\tan ^2 1+\cot ^2 1=\frac{\sin ^4 1+\cos ^4 1}{\sin ^2 1 \cos ^2 1}$
$\Rightarrow a=\frac{\left(\sin ^2 1+\cos ^2 1\right)^2-2 \sin ^2 1 \cot ^2 1}{(\sin 1 \cos 1)^2}$
$\Rightarrow a=\frac{1-\frac{1}{2} \sin ^2 2}{\frac{1}{4} \sin ^2 2}=\frac{4-2 \sin ^2 2}{\sin ^2 2}=\frac{4-(1-\cos 4)}{\frac{1}{2}(1-\cos 4)}$
$\Rightarrow a=\frac{2(3+\cos 4)}{1-\cos 4}$