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 find:

• The differential $dx = 8 \sin \theta \cos \theta d\theta$.
• Limits change from $x=0$ (when $\theta = 0$) to $x=3$ (which gives $\theta = \frac{\pi}{6}$).

We rewrite the integral:

$\int_0^{3} \frac{x}{\sqrt{4 - x}} dx = \int_0^{\frac{\pi}{6}} \frac{4 \sin^2 \theta}{\sqrt{4 - 4 \sin^2 \theta}} (8 \sin \theta \cos \theta) d\theta$

Since $\sqrt{4(1 - \sin^2 \theta)} = 2 \cos \theta$, this becomes:

$= \int_0^{\frac{\pi}{6}} \frac{4 \sin^2 \theta \cdot 8 \sin \theta \cos \theta}{2 \cos \theta} d\theta$ $= 16 \int_0^{\frac{\pi}{6}} \sin^3 \theta d\theta$

This integral can be evaluated using the reduction formula or integration by parts, leading to the final identity involving $\lambda = 16$.

Now applying the earlier results, we can use the fundamental theorem of calculus and the integration we have conducted.

Given:

• The integral equals $\lambda \int_0^{\frac{\pi}{2}} \sin^2 \theta d\theta$, where we know the integral value is half, contributing to:

$\int \sin^2 \theta d\theta = \frac{\theta}{2} - \frac{1}{4} \sin(2\theta)$ Integrating $0$ to $\frac{\pi}{2}$ yields:

$= \frac{\frac{\pi}{2}}{2} = \frac{\pi}{4}$ Thus, $\lambda \frac{\pi}{4} = \frac{16\pi}{4} = 4\pi$

The final expression can then be substituted back to form: $\int \frac{x}{\sqrt{4 - x}} dx = 4\pi + C$

In conclusion, in the form $ax + b$, where $a = 4$ and $b = \pi$.