Problem source

\begin{questionparts}
\item Let
$$f(x) = \frac{x}{\sqrt{x^2 + p}},$$
where $p$ is a non-zero constant. Sketch the curve $y = f(x)$ for $x \geq 0$ in the case $p > 0$.
\item Let
$$I = \int \frac{1}{(b^2 - y^2)\sqrt{c^2 - y^2}} \, dy,$$
where $b$ and $c$ are positive constants. Use the substitution $y = \frac{cx}{\sqrt{x^2 + p}}$, where $p$ is a suitably chosen constant, to show that
$$I = \int \frac{1}{b^2 + (b^2 - c^2)x^2} \, dx.$$
Evaluate
$$\int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \, dy.$$
[ Note: $\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + \text{constant.}$ ]
Hence evaluate
$$\int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \, dy.$$
\item By means of a suitable substitution, evaluate
$$\int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \, dy.$$
\end{questionparts}

Solution source

\begin{questionparts}

\item $\,$ 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)/sqrt((#1)^2+4};
    \def\xl{-1};
    \def\xu{8};
    \def\yl{-.3};
    \def\yu{1.5};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);

        % \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
        % \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
        % \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
        
        \draw[thick, blue, smooth, domain=0:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

        \draw[dashed, red] (\xl, 1) -- (\xu, 1) node[pos=0.5, above] {$y = 1$};

        \node[blue, above, rotate=60] at (1, {\functionf(1)}) {\tiny $y = \frac{x}{\sqrt{x^2+p}}$}; 
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}


\item $\,$ \begin{align*}
&& y &= \frac{cx}{\sqrt{x^2+p}} \\
&& \d y &= \frac{c(x^2+p)-cx^2}{(x^2+p)^{3/2}} \d x \\
&&&= \frac{cp^2}{(x^2+p)^{3/2}} \d x\\
&& I &= \int \frac1{(b^2-y^2)\sqrt{c^2-y^2}} \d y \\
&&&= \int \frac{1}{\left ( b^2 - \frac{c^2x^2}{x^2+p} \right) \sqrt{c^2 - \frac{c^2x^2}{x^2+p}  }} \d y \\
&&&= \int \frac{(x^2+p)^{3/2}}{((b^2-c^2)x^2+pb^2)\sqrt{c^2p}}\frac{cp}{(x^2+p)^{3/2}} \d x \\
&&&= \int \frac{\sqrt{p}}{((b^2-c^2)x^2+pb^2)} \d x \\
p=1: &&&= \int \frac{1}{(b^2-c^2)x^2+b^2} \d x 
\end{align*}
When $b = \sqrt{3}, c = \sqrt{2}$
\begin{align*}
&& I_1 &= \int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \d y\\
&&&= \int_{x =1 }^{x=\infty} \frac{1}{3+x^2} \d x \\
&&&= \left [ \frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} \right]_1^\infty \\
&&&= \frac{\pi}{2\sqrt{3}} - \frac{1}{\sqrt{3}} \frac{\pi}{6} \\
&&&= \frac{\pi}{3\sqrt{3}}
\end{align*}

\begin{align*}
&& I_2 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\
x = \frac1y, \d x = -\frac1{y^2} \d y &&&= \int_{x=\sqrt{2}}^{x=1} \frac{x^2}{(3-x^2)\sqrt{2-x^2}}\cdot \left ( -\frac{1}{x^2} \right ) \d x \\
&&&=  \int_1^{\sqrt{2}} \frac{1}{(3-x^2)\sqrt{2-x^2}} \d x \\
&&&= I_1 = \frac{\pi}{3\sqrt{3}}
\end{align*}

\item $\,$ \begin{align*}
&& I_3 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\
x = 1/y, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{2}} \frac{x}{(3-x^2)\sqrt{2-x^2}} \d x \\
u = x^2, \d u = 2x \d x &&&= \int_{u=1}^{u=2}  \frac{\frac12}{(3-u)\sqrt{2-u}} \d u \\
v=2-u, \d v = -\d u &&&= \frac12\int_{v=0}^{v=1} \frac{1}{(1+v)\sqrt{v}} \d v \\
&&&=\left [\tan^{-1}\sqrt{v}\right]_0^1 \\
&&&= \frac{\pi}{4}
\end{align*}

\end{questionparts}