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# Help with Edexcel P5 Questions watch

1. Hi guys just a quick one that's probably glaringly obvious. It's from the P5 specimen paper Q6.

I have:

a∫√(2-2cost) dt.

I have racked my brains for identities to use but am stuck. The answer scheme comes up with 2a∫sin(0.5t) dt. Why is this? Could someone explain please.

Cheers, Ian.
2. (Original post by The_Hunter)
Hi guys just a quick one that's probably glaringly obvious. It's from the P5 specimen paper Q6.

I have:

a∫√(2-2cost) dt.

I have racked my brains for identities to use but am stuck. The answer scheme comes up with 2a∫sin(0.5t) dt. Why is this? Could someone explain please.

Cheers, Ian.
a∫(2-2cost)^0.5 dt
=a∫[2-2(1-2(sin0.5t)^2]^0.5 dt
=a∫[4(sin0.5t)^2] dt
=a∫2sin0.5t dt
=2a∫sin(0.5t) dt

[Cos2A=(cosA)^2-(sinA)^2=2(cosA)^2-1=1-2(sinA)^2]
3. (Original post by Gaz031)
a∫(2-2cost)^0.5 dt
=a∫[2-2(1-2(sin0.5t)^2]^0.5 dt
=a∫[4(sin0.5t)^2] dt
=a∫2sin0.5t dt
=2a∫sin(0.5t) dt

[Cos2A=(cosA)^2-(sinA)^2=2(cosA)^2-1=1-2(sinA)^2]
Not too difficult really was it! Thanks for your help.

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