Integration help pls

13

Hi,
Pls could I have some help integrating this. I’ve tried it below but I didn’t get the correct answer, I’m not sure where i went wrong though.
Here is the my attempt: https://ibb.co/hf1JYj6
Thanks!

Reply 1

mqb2766

21

Original post

by anonymous56754

Hi,
Pls could I have some help integrating this. I’ve tried it below but I didn’t get the correct answer, I’m not sure where i went wrong though.
Here is the my attempt: https://ibb.co/hf1JYj6
Thanks!

Did you have any reason for using the pythagorean identity at the start? Was there any other thing you could do or ....?

But I suspect the error is your (incorrect) identity for cos^4(x). If youre going down this route, maybe use the fact its (cos^2(x))^2?

Reply 2

davros

16

Original post

by anonymous56754

Hi,
Pls could I have some help integrating this. I’ve tried it below but I didn’t get the correct answer, I’m not sure where i went wrong though.
Here is the my attempt: https://ibb.co/hf1JYj6
Thanks!

your replacement of cos^4x with an expression in cos4x is incorrect - you're ok going from cos^2 x to cos2x but remember that cos^4 x = (cos^2 x)^2 so you can't just use the wrong expression you've put in!

As an alternative method, you could write sin2xcos2x = (1/2)sin4x initially and see where that leads :smile:

Reply 3

anonymous56754

OP

13

ah, thank you everyone. I think replacing cos^4x with (cos^2x)^2 is too long so I tried a different method and it worked!

Reply 4

mqb2766

21

Original post

by anonymous56754

ah, thank you everyone. I think replacing cos^4x with (cos^2x)^2 is too long so I tried a different method and it worked!

Presuming you went with
integrand ~ sin^2(2x) ~ 1 + cos(4x)?
When you start off a problem like this, its best to try and consider all possible (sensible) identities and think a couple of lines ahead about which ones are likely to give. The values of the constants etc are not that important when thinking about how to start.

Here you want to get to something llike
trig(4x)
as you have the product of two squared trig terms so roughly a product of four trig(x). So subbing the squared/product terms using either cos(2x) or sin(2x) or pythagoras are the main contenders, then repeat as youd still have a product of two trig(2x) terms to get the trig(4x) term. Eventually you should get there with any correct approach, though you may end up making too many mistakes or cycling or ...

Reply 5

anonymous56754

OP

13

Original post

by mqb2766

Presuming you went with
integrand ~ sin^2(2x) ~ 1 + cos(4x)?
When you start off a problem like this, its best to try and consider all possible (sensible) identities and think a couple of lines ahead about which ones are likely to give. The values of the constants etc are not that important when thinking about how to start.
Here you want to get to something llike
trig(4x)
as you have the product of two squared trig terms so roughly a product of four trig(x). So subbing the squared/product terms using either cos(2x) or sin(2x) or pythagoras are the main contenders, then repeat as youd still have a product of two trig(2x) terms to get the trig(4x) term. Eventually you should get there with any correct approach, though you may end up making too many mistakes or cycling or ...

I went with converting sin^2x to (1/2-1/2cos2x) and converted cos^2x too then expanded, I just found it easier for me 🙂

Reply 6

mqb2766

21

Original post

by anonymous56754

I went with converting sin^2x to (1/2-1/2cos2x) and converted cos^2x too then expanded, I just found it easier for me 🙂

Half of this stuff is practice and trying different ways. Personally (not worrying about the exact values of the constants), Id have recognised that
2sin(x)cos(x) = sin(2x)
so squaring
4sin^2(x)cos^2(x) = sin^2(2x)
then used the double angle for sin(2x) so
2sin^2(2x) = 1 - cos(4x)
So the right hand side is (roughly) the integrand and its ~2 lines of working. But with identities, there are several routes you can often take.

Reply 7

anonymous56754

OP

13

Original post

by mqb2766

Half of this stuff is practice and trying different ways. Personally (not worrying about the exact values of the constants), Id have recognised that
2sin(x)cos(x) = sin(2x)
so squaring
4sin^2(x)cos^2(x) = sin^2(2x)
then used the double angle for sin(2x) so
2sin^2(2x) = 1 - cos(4x)
So the right hand side is (roughly) the integrand and its ~2 lines of working. But with identities, there are several routes you can often take.

then used the double angle for sin(2x) so 2sin^2(2x) = 1 - cos(4x)

sorry but for this step, isnt sin(2x)= 2sin(x)cos(x)

Reply 8

mqb2766

21

>> then used the double angle for sin(2x) so 2sin^2(2x) = 1 - cos(4x)
>> sorry but for this step, isnt sin(2x)= 2sin(x)cos(x)

Isnt the bold what I wrote (and similarly for the first identity?)?

Is the question ... There are (at least) two "double angle" identities involving sin(2x) so
sin(2x) = 2sin(x)cos(x)
cos(4x) = 1 - 2sin^2(2x)
or cos(2A) = 1 - sin^2(A) where A=2x. tts probably the simplest way to do this question. Its ok to do it other ways, but its a problem if you dont really understand the simple way. I guess Im abusing terminology somewhat to call the second one a double angle for sin(2x) as that usually means
sin(2x) = some function of trig(x)
whereas for the second I mean
sin(2x) = some function of trig(4x)
but the original reply about thinking going
trig^4(x) -> trig^2(2x) -> trig(4x)
stands.

(edited 10 months ago)

Integration help pls

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