Use the substitution $t = \tan \frac{\theta}{2}$ evaluate $$\int_0^{\frac{\pi}{2}} \frac{d\theta}{2 - \cos \theta}$$ A falling particle experiences forces due to gravity and air resistance - HSC - SSCE Mathematics Extension 2 - Question 14 - 2018 - Paper 1

Using the equation of motion, we can relate acceleration and velocity:

$a = \frac{dv}{dt} = -kv^2$

By separating variables and integrating, we find:

$\int \frac{1}{v^2} \, dv = -k \int dt \\ v = \frac{1}{C - kt}$

Using the condition that the particle falls from rest allows us to solve for $C$, leading to:

$v = \sqrt{\frac{1}{k}(1 - e^{-2kh})}$

Using integration by parts, we can express the integral as:

$I_n = \int_{-3}^0 \sqrt{x^3 + 5} \, dx = \frac{1}{n} \cdots \text{ (similar calculations)}$

Through a series of transformations, you arrive at the equation:\

$I_n = \frac{-6n}{3 + 2n} I_{n - 1}$

To find $I_2$, substitute $n = 2$ into the earlier derived formula:

$I_2 = \frac{-12}{7} I_1$

Evaluating $I_1$ and then substituting back will yield the value of $I_2$.

The probability that player A wins every game can be calculated using:

$P(A \text{ wins all games}) = \left( \frac{1}{3} \right)^n$

This can be derived from the total probability principle:

For $n \geq 2$, evaluating gives:

$P(A \text{ wins}) \cdot P(B \text{ wins}) \cdots$

This will lead to:

$\left(\frac{2}{3}\right)^{n - 2} \left(\frac{1}{3}\right)$

Applying the principle of inclusion-exclusion yields:

$P(A \cup B \cup C) = 1 - P(A' \cap B' \cap C')$

Calculating gives:

$\frac{3^{n - 1} - 2^{n} + 1}{3^{n - 1}}$. This concludes the proof.