User:Xenocidius/Specialist/Q3

i.
$$ \begin{align} f'(x) & = \frac{d}{dx} ( \sin^{-1} \frac{x-1}{x+1} ) \\ & = \frac{d}{dx} ( \frac{x-1}{x+1} ) \frac{1}{\sqrt{1 - (\frac{x-1}{x+1})^2}} \\ & = \frac{d}{dx} ( 1 - \frac{2}{x+1} ) \frac{1}{\sqrt{1 - \frac{(x-1)^2}{(x+1)^2}}} \\ & = \frac{2}{(x+1)^2} \frac{1}{\sqrt{1 - \frac{(x-1)^2}{(x+1)^2}}} \\ & = \frac{2}{\sqrt{(x+1)^4}} \frac{1}{\sqrt{1 - \frac{(x-1)^2}{(x+1)^2}}} \\ & = \frac{2}{\sqrt{(x+1)^4 - (x-1)^2(x+1)^2}} \\ & = \frac{2}{(x+1)\sqrt{(x+1)^2 - (x-1)^2}} \\ & = \frac{2}{(x+1)\sqrt{x^2 + 2x + 1 - (x^2 - 2x + 1))}} \\ & = \frac{2}{(x+1)\sqrt{4x}} \\ & = \frac{1}{\sqrt{x}(x+1)} \end{align} $$

ii.
$$ \begin{align} g'(x) & = \frac{d}{dx} ( 2 \tan^{-1} \sqrt{x} ) \\ & = \frac{1}{2\sqrt{x}} \frac{2}{1 + \sqrt{x}^2} \\ & = \frac{1}{\sqrt{x}(x + 1)} \end{align} $$

iii.
$$ \begin{align} f(x) - g(x) & = \int \! f'(x) - g'(x) \,dx \\ & = \int \! \frac{1}{\sqrt{x}(x + 1)} - \frac{1}{\sqrt{x}(x + 1)} \,dx \\ & = \int \! 0 \,dx \\ & = c \\

f(1) - g(1) & = \sin^{-1} \frac{1-1}{1+1} - 2 \tan^{-1} \sqrt{1} \\ & = \sin^{-1} 0 - 2 \tan^{-1} 1 \\ & = 0 - 2 \frac{\pi}{4} \\ & = - \frac{\pi}{2} \\

\therefore c & = - \frac{\pi}{2} \\ \therefore f(x) - g(x) & = - \frac{\pi}{2} \end{align} $$

$$ \begin{align} f(0) - g(0) & = \sin^{-1} \frac{0-1}{0+1} - 2 \tan^{-1} \sqrt{0} \\ & = \sin^{-1} -1 - 2 \tan^{-1} 0 \\ & = - \frac{\pi}{2} \\ \end{align} $$

iv.
$$ \begin{align} & \int \! \frac{1}{\sqrt{x}(x + 1)} \,dx = 2 \tan^{-1} \sqrt{x} \\ \therefore & \int \! \frac{1}{x^{3/2} + x^{1/2}} \,dx = 2 \tan^{-1} \sqrt{x} \\ \therefore & \int \! \frac{6}{x^{3/2} + x^{1/2}} \,dx = 12\tan^{-1} \sqrt{x} \\ \therefore & \int_1^3 \! \frac{6}{x^{3/2} + x^{1/2}} \,dx = 12\tan^{-1} \sqrt{3} - 12\tan^{-1} \sqrt{1} \\ & = 12 \frac{\pi}{3} - 12\frac{\pi}{4} \\ & = 4\pi - 3\pi \\ & = \pi \end{align} $$