Prove the identity $$\frac{\sin 2x}{1 + \tan^2 x} = 2 \sin x \cos^3 x$$ [3 marks] Hence find $$\int \frac{4 \sin 4\theta}{1 + \tan^2 2\theta} d\theta$$ [6 marks] - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 3

To find the integral:

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

we first rewrite the integral using the identity established:

Here, we know:

$1 + \tan^2 2\theta = \sec^2 2\theta$

Thus,

$\int \frac{4 \sin 4\theta}{\sec^2 2\theta} d\theta = \int 4 \sin 4\theta \cos^2 2\theta \, d\theta$

Next, we can express (\sin 4\theta) in terms of (\sin 2\theta):

$\sin 4\theta = 2 \sin 2\theta \cos 2\theta$

Substituting this gives:

$\int 4 (2 \sin 2\theta \cos 2\theta) \cos^2 2\theta \, d\theta$

Rearranging and simplifying leads to:

$\int 8 \sin 2\theta \cos^3 2\theta \, d\theta$

Now we can use the substitution method:

Let (u = \sin 2\theta) so that (du = 2\cos 2\theta , d\theta$$

We obtain:

$\int 4u \cos^2 2\theta \, du$

Now substituting and solving yields:

Finally, the specific integration leads to a result of:

$= -4 \cos^2 2\theta \cdot \sin 2\theta + C$

This is our final answer.