By using the substitution $x = a\tan\theta$, show that $\displaystyle \int \frac{1}{x\sqrt{x^2 + a^2}}\,dx = \frac{1}{a}\ln\frac{x}{\sqrt{x^2 + a^2} + a} + c ,$ where $x > 0$ and $a > 0$. At time $t$ in a chemical reaction, $x$ kg of substance X and $y$ kg of substance Y are present. Initially, there is 4 kg of X and 5 kg of Y. The variables x and y satisfy the equations $\dfrac{dx}{dt} = xy$ and $\dfrac{dy}{dt} = x^2$. Find $\dfrac{dy}{dx}$ in terms of x and y, and by solving this differential equation, show that $y = \sqrt{x^2 + 9}.$ Obtain a differential equation in terms of $x$ and $t$ only and hence find an expression of $t$ in terms of $x$.

1. Find the differential $dx$ in terms of $d\theta$.

Given $x = a\tan\theta$,
$\frac{dx}{d\theta} = a\sec^2\theta \implies dx = a\sec^2\theta \, d\theta$

2. Substitute $x$ and $dx$ into the integral and simplify the expression inside the square root.

$\int \frac{1}{x\sqrt{x^2 + a^2}}\,dx = \int \frac{1}{a\tan\theta\sqrt{a^2\tan^2\theta + a^2}} (a\sec^2\theta) \, d\theta$
$= \int \frac{a\sec^2\theta}{a\tan\theta\sqrt{a^2(\tan^2\theta + 1)}} \, d\theta$

3. Simplify the trigonometric integrand using the identity $1 + \tan^2\theta = \sec^2\theta$.

Since $\tan^2\theta + 1 = \sec^2\theta$ and for $x>0$ and $a>0$, $\theta$ is in the first quadrant, so $\sec\theta > 0$. The square root simplifies to $a\sec\theta$.
$= \int \frac{a\sec^2\theta}{(a\tan\theta)(a\sec\theta)} \, d\theta$
$= \int \frac{\sec\theta}{a\tan\theta} \, d\theta$
$= \frac{1}{a} \int \frac{1/\cos\theta}{\sin\theta/\cos\theta} \, d\theta$
$= \frac{1}{a} \int \csc\theta \, d\theta$

4. Integrate with respect to $\theta$ using the standard integration formula.

Using the standard formula from the List of Formulae (MF26),
$= -\frac{1}{a}\ln|\csc\theta + \cot\theta| + c$

5. Express $\csc\theta$ and $\cot\theta$ in terms of $x$ and $a$.

From $x = a\tan\theta$, we have $\tan\theta = \frac{x}{a}$. By drawing a right-angled triangle with opposite side $x$ and adjacent side $a$, the hypotenuse is $\sqrt{x^2 + a^2}$.
$\csc\theta = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{\sqrt{x^2 + a^2}}{x}$
$\cot\theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{a}{x}$

6. Substitute back into the expression and simplify using logarithm properties to match the target form.

$= -\frac{1}{a}\ln\left| \frac{\sqrt{x^2 + a^2}}{x} + \frac{a}{x} \right| + c$
$= -\frac{1}{a}\ln\left| \frac{\sqrt{x^2 + a^2} + a}{x} \right| + c$
Using the logarithm property $- \ln(A) = \ln(A^{-1})$, we invert the fraction inside the logarithm:
$= \frac{1}{a}\ln\left| \frac{x}{\sqrt{x^2 + a^2} + a} \right| + c$
Since $x > 0$ and $a > 0$, the expression inside the modulus is positive, allowing us to drop the absolute value signs:
$= \frac{1}{a}\ln\left( \frac{x}{\sqrt{x^2 + a^2} + a} \right) + c \quad \text{(shown)}$

1. Find $\frac{dy}{dx}$ using the given rate equations.

By the chain rule,
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{x^2}{xy} = \frac{x}{y}$

2. Separate the variables and integrate both sides of the differential equation.

$\frac{dy}{dx} = \frac{x}{y}$
$y \, dy = x \, dx$
$\int y \, dy = \int x \, dx$
$\frac{1}{2}y^2 = \frac{1}{2}x^2 + C$

3. Apply the initial conditions to find the constant of integration.

Initially (at $t=0$), $x = 4$ and $y = 5$.
$\frac{1}{2}(5)^2 = \frac{1}{2}(4)^2 + C$
$\frac{25}{2} = 8 + C$
$C = \frac{9}{2}$

4. Rearrange to make $y$ the subject and justify the chosen sign for the square root.

$\frac{1}{2}y^2 = \frac{1}{2}x^2 + \frac{9}{2}$
$y^2 = x^2 + 9$
$y = \pm\sqrt{x^2 + 9}$
Since $y$ represents the mass of a substance, $y > 0$, so we reject the negative root.
$y = \sqrt{x^2 + 9} \quad \text{(shown)}$

1. Substitute $y$ in terms of $x$ to obtain a differential equation in $x$ and $t$.

$\frac{dx}{dt} = xy$
Substitute $y = \sqrt{x^2 + 9}$ (from part b(i)):
$\frac{dx}{dt} = x\sqrt{x^2 + 9}$

2. Separate the variables to set up the integral.

$\int 1 \, dt = \int \frac{1}{x\sqrt{x^2 + 9}} \, dx$

3. Integrate both sides using the given identity from part (a).

Comparing the integral with part (a), we use $a = 3$:
$t = \frac{1}{3}\ln\left(\frac{x}{\sqrt{x^2 + 9} + 3}\right) + C$

4. Use initial conditions to find the constant of integration and simplify the final expression.

Initially, $t = 0$ and $x = 4$:
$0 = \frac{1}{3}\ln\left(\frac{4}{\sqrt{4^2 + 9} + 3}\right) + C$
$0 = \frac{1}{3}\ln\left(\frac{4}{5 + 3}\right) + C$
$0 = \frac{1}{3}\ln\left(\frac{1}{2}\right) + C$
$C = -\frac{1}{3}\ln\left(\frac{1}{2}\right) = \frac{1}{3}\ln 2$
Substitute $C$ back into the equation and use logarithm properties to combine terms:
$t = \frac{1}{3}\ln\left(\frac{x}{\sqrt{x^2 + 9} + 3}\right) + \frac{1}{3}\ln 2$
$t = \frac{1}{3}\ln\left(\frac{2x}{\sqrt{x^2 + 9} + 3}\right)$