velaris08
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when integrating cos^2 (2x-1), i used the identity cos 2A = 2cos^2A - 1, manipulated it to (1/2)cos (4x - 2) and equalled it - which was the wrong approach. can you not equal them because A has to be like strictly a product of two things? like can it not be i.e. 2x - 1?
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mqb2766
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(Original post by velaris08)
when integrating cos^2 (2x-1), i used the identity cos 2A = 2cos^2A - 1, manipulated it to (1/2)cos (4x - 2) and equalled it - which was the wrong approach. can you not equal them because A has to be like strictly a product of two things? like can it not be i.e. 2x - 1?


cos(2A) = 2cos^2(A) - 1
So


cos^2(A) = cos(2A)/2 + 1/2
and integrate each term when A=2x-1. It helps to see your actual working, you've got the right idea.
Last edited by mqb2766; 1 month ago
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velaris08
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(Original post by mqb2766)


cos(2A) = 2cos^2(A) - 1
So


cos^2(A) = cos(2A)/2 + 1/2
and integrate each term when A=2x-1. It helps to see your actual working, you've got the right idea.
cos 2A / 2 + 1 / 2 ? Where’s the first 2 that’s not 2A coming from? Attaching a picture of my working. It’s not that I don’t get what sol bank says - rather that in the exam I would probably have approached it my way and I don’t get what mistake I’ve made.
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mqb2766
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(Original post by velaris08)
cos 2A / 2 + 1 / 2 ? Where’s the first 2 that’s not 2A coming from? Attaching a picture of my working. It’s not that I don’t get what sol bank says - rather that in the exam I would probably have approached it my way and I don’t get what mistake I’ve made.
Your working looks fine and what you've written is the same as in the previous post. Note it helps to put the arguments of the trig function in brackets, so its clear whether operations apply to the argument or to the trig function. If the solution bank says different, can you pls upload it. If you're asking about something else, can you pls clearly state it.
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velaris08
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(Original post by mqb2766)
Your working looks fine and what you've written is the same as in the previous post. Note it helps to put the arguments of the trig function in brackets, so its clear whether operations apply to the argument or to the trig function. If the solution bank says different, can you pls upload it. If you're asking about something else, can you pls clearly state it.
Here’s what sol bank says!
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mqb2766
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(Original post by velaris08)
Here’s what sol bank says!
Can you upload a picture of the original question. It looks like theyre doing
(cos(2x)-1)^2
not
cos^2(2x-1)
Last edited by mqb2766; 1 month ago
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velaris08
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(Original post by mqb2766)
Can you upload a picture of the original question. It looks like theyre doing
(cos(2x)-1)^2
not
cos^2(2x-1)
“integrate the following” - oh god just realised you’re right, it’s the brackets thing! i mentally added them im so sorry
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mqb2766
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(Original post by velaris08)
“integrate the following” - oh god just realised you’re right, it’s the brackets thing! i mentally added them im so sorry
Arguably the original question should have been clearer, but its not unusual to write it like that.
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