Edexcel Pure Paper 1 2024 Question 13

Part (a)

We are given the substitution:

x = a\sin^2\theta

Differentiate:

\frac{dx}{d\theta} = 2a\sin\theta\cos\theta

Now substitute into the integral:

\int_0^a x^{1/2}\sqrt{a – x},dx

Replace each part carefully.

First:

x^{1/2} = \sqrt{a\sin^2\theta} = \sqrt{a}\sin\theta

Second:

\sqrt{a – x} = \sqrt{a – a\sin^2\theta} = \sqrt{a(1 – \sin^2\theta)} = \sqrt{a}\cos\theta

Now include dx:

dx = 2a\sin\theta\cos\theta,d\theta

Putting everything together:

\int \sqrt{a}\sin\theta \cdot \sqrt{a}\cos\theta \cdot 2a\sin\theta\cos\theta, d\theta

Simplify:

= 2a^2 \int \sin^2\theta \cos^2\theta , d\theta

Now adjust limits.

When x = 0:

\sin^2\theta = 0 \Rightarrow \theta = 0

When x = a:

\sin^2\theta = 1 \Rightarrow \theta = \frac{\pi}{2}

So:

= 2a^2 \int_0^{\frac{\pi}{2}} \sin^2\theta \cos^2\theta , d\theta

Use identity:

\sin^2\theta \cos^2\theta = \left(\frac{1}{2}\sin 2\theta\right)^2 = \frac{1}{4}\sin^2 2\theta

So:

= \frac{1}{2}a^2 \int_0^{\frac{\pi}{2}} \sin^2 2\theta , d\theta

✅ Result (a)

\frac{1}{2}a^2 \int_0^{\frac{\pi}{2}} \sin^2 2\theta , d\theta