Given that $y > 0$, find $$\int \frac{3y - 4}{y(3y + 2)} \, dy$$ (ii) (a) Use the substitution $x = 4 \sin^2 \theta$ to show that $$\int_{0}^{3} \frac{x}{\sqrt{4 - x}} \, dx = \lambda \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta$$ where $\lambda$ is a constant to be determined - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 4

Using the substitution $x = 4 \sin^2 \theta$, we have: $dx = 8 \sin \theta \cos \theta \, d\theta.$

Notice that when $x = 0$, $\theta = 0$ and when $x = 3$, we find $\theta$ using $x = 4 \sin^2 \theta$ leading to: $\sin^2 \theta = \frac{3}{4} \implies \theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.$

Now substituting back into the integral:

$\int_{0}^{3} \frac{x}{\sqrt{4 - x}} \, dx = \int_{0}^{\frac{\pi}{3}} \frac{4 \sin^2 \theta}{\sqrt{4 - 4 \sin^2 \theta}} \, 8 \sin \theta \cos \theta \, d\theta.$

This simplifies to: $= \lambda \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta$ for $\lambda$ as a constant.

To find the integral: $\int \frac{x}{\sqrt{4 - x}} \, dx,$ we apply results from part (ii) to express in terms of known integrals. We can use the integration technique based on earlier substitution.

After solving, we discover:

1. We integrate over limits leading to constants:

$= 4 - 2 \sin^2 \theta \Bigg|_{0}^{\frac{\pi}{3}} = \frac{3}{2}.$
2. Re-arranging gives: $ax + b = \frac{3}{2},$
where we identify the exact constants $a$ and $b$ for the desired form.