Trig identities in integration question.

[thiago1]

Hi I'm stuck in an integration question where you have to do a lot of simplifying before integrating, and one of the steps is beyond me. Here's the question for context;

Here's the mark scheme with a line under the part that I'm unsure on.

I don't know how to go from 1/4sin^2(2θ) to 1/8(1-cos4θ). Thanks in advance.

 

[lex]

Use the double angle formula:
[MATH]\cos 2x = 1-2\sin^2x\\ \text{with }x \text{ being } 2\theta[/MATH]

 

[nasi112]

[MATH]\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}[/MATH]

 

[Steven G]

I don't know how to go from 1/4sin^2(2θ) to 1/8(1-cos4θ). Thanks in advance.

1/4sin^2(2θ) to 1/8(1-cos4θ)
Multiply both sides by 4, get sin^2(2θ) to 1/2(1-cos4θ) ==>sin^2(θ) to 1/2(1-cos2θ) and call this an identity!