8 (a) Prove the identity $$\frac{\sin 2x}{1 + \tan^2 x} = 2 \sin x \cos^3 x$$ 8 (b) Hence find $$\int \frac{4 \sin 4\theta}{1 + \tan^2 2\theta} d\theta$$ - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 3

To solve the given integral, we first rewrite the integrand:

We can use the identity for (\tan^2) to express the integral as follows:

$\int \frac{4 \sin 4\theta}{1 + \tan^2 2\theta} d\theta = \int 4 \sin 4\theta \cos^2 2\theta d\theta$

Let (u = \cos 2\theta), then:\n(du = -2 \sin 2\theta d\theta) or (d\theta = -\frac{du}{2 \sin 2\theta}).

Substituting:

$= \int -2 \sin 4\theta u^2 d\theta$

Rewriting (\sin 4\theta = 2 \sin 2\theta \cos 2\theta = -2 \sin 2\theta (1-u^2)), we get:

$= \int -4u^2 (-\frac{du}{2(1-u^2)})$

Solving this integral leads to the final answer being of the form:

$= -\frac{1}{3}u^3 + C$

Replacing back for u:

$= -\frac{1}{3}\cos^3 2\theta + C$.