# integrate sec^4 x dxWatch

This discussion is closed.
#1
plz can u integrate sec^4 xdx thanks!
0
14 years ago
#2
(Original post by tammypotato)
plz can u integrate sec^4 xdx thanks!
Yeah plzzzzzzzzz I've been stuck on it for ages
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14 years ago
#3
(Original post by Pixelfairy #1)
Yeah plzzzzzzzzz I've been stuck on it for ages
sec^4x = sec^2x(1 + tan^2x)

= sec^2x + sec^2xtan^2x

=> INT = tanx + (tan^3x)/3 + C
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#4
(Original post by It'sPhil...)
sec^4x = sec^2x(1 + tan^2x)

= sec^2x + sec^2xtan^2x

=> INT = tanx + (tan^3x)/3 + C
lol thank u soooo much
0
14 years ago
#5
(Original post by It'sPhil...)
sec^4x = sec^2x(1 + tan^2x)

= sec^2x + sec^2xtan^2x

=> INT = tanx + (tan^3x)/3 + C
Is there a trick to doing questions like that, or did you just learn it?
0
14 years ago
#6
(Original post by Pixelfairy #1)
Is there a trick to doing questions like that, or did you just learn it?
I think it's practice...I'm fed up with calculus...I've been doing P5 for two hours now.
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14 years ago
#7
(Original post by Squishy)
I think it's practice...I'm fed up with calculus...I've been doing P5 for two hours now.
Are you not nearly suicidal
0
14 years ago
#8
(Original post by Pixelfairy #1)
Is there a trick to doing questions like that, or did you just learn it?
If you have highish powers of trigonometry you usually want to break it down into a product of two functions. In general if you vave a power of sec or tan youre gonna have to use tan^2x + 1 = sec^2x and d/dx(tanx) = sec^2x and d/dx(secx) = secxtanx. So not only are sec and tan related by trig they are also related through calculus. But practice is important, as you pick up the kinds of techniques to use
0
14 years ago
#9
(Original post by Pixelfairy #1)
Are you not nearly suicidal
Nope, it's just that I didn't revise for my AS maths modules, so I thought I'd do it for my A2s, since these grades actually mean something to me this time.
0
14 years ago
#10
(Original post by Pixelfairy #1)
Are you not nearly suicidal
P5 is the most interesting module I have studied; the questions are slightly less routine and are a bit more difficult. However, 2 hours of it would probably get fairly boring.
0
14 years ago
#11
(Original post by It'sPhil...)
In general if you vave a power of sec or tan youre gonna have to use tan^2x + 1 = sec^2x and d/dx(tanx) = sec^2x and d/dx(secx) = secxtanx.
Yeah, that's good advice. Sec/tan and cosec/cot are almost always related in the questions they give you.
0
14 years ago
#12
(Original post by mikesgt2)
P5 is the most interesting module I have studied; the questions are slightly less routine and are a bit more difficult. However, 2 hours of it would probably get fairly boring.
When I offered to self-study Further Maths, P5 was the module they warned me about. It is much less standard than P6. It almost feels like a non-sequitur between P4 and P6. 2 hours is the longest I can do maths without a break...good thing, since my exams are an hour and a half.
0
14 years ago
#13
(Original post by It'sPhil...)
If you have highish powers of trigonometry you usually want to break it down into a product of two functions. In general if you vave a power of sec or tan youre gonna have to use tan^2x + 1 = sec^2x and d/dx(tanx) = sec^2x and d/dx(secx) = secxtanx. So not only are sec and tan related by trig they are also related through calculus. But practice is important, as you pick up the kinds of techniques to use
Yeah, integration is often about learning tricks and gaining experience: I would recommend just sitting down and doing loads of integration from the textbook, you get to learn all the required methods and things such as substitution and parts become second nature. The same applies for differentiation, it will really help in the exam if you can just do calculus like second nature and apply all the chain, product, quotient rules without too much thought. It is all about practice and often in maths exams it is not about whether you are a genius, it is about how many of that particular class of questions you have done before and how familiar you are with the methods. Perhaps that is just a cynical view of maths exams, but I think that lots of practice is the way to go.
0
14 years ago
#14
(Original post by mikesgt2)
Yeah, integration is often about learning tricks and gaining experience: I would recommend just sitting down and doing loads of integration from the textbook, you get to learn all the required methods and things such as substitution and parts become second nature. The same applies for differentiation, it will really help in the exam if you can just do calculus like second nature and apply all the chain, product, quotient rules without too much thought. It is all about practice and often in maths exams it is not about whether you are a genius, it is about how many of that particular class of questions you have done before and how familiar you are with the methods. Perhaps that is just a cynical view of maths exams, but I think that lots of practice is the way to go.
Thats certainly what ive found for A level exams - there are a few questions which require a bit of insight but in general it is about how accurately you can carry out the methods that you have learnt. However if you do STEP papers you are constantly thinking of new ideas - nothing is ever obvious, you are rarely told what to do and you have to think about things in a very abstract way before getting the answer. Often you can do a solution in a side or so which is about the same as a longish P5/P6 question but getting to the solution is a much more challenging and rewarding experience. Are you doing any STEP Mike?
0
14 years ago
#15
(Original post by It'sPhil...)
sec^4x = sec^2x(1 + tan^2x)

= sec^2x + sec^2xtan^2x

=> INT = tanx + (tan^3x)/3 + C
how does sec^2x.tan^2x become (tan^3x/3)???? thx
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14 years ago
#16
(Original post by wonkey)
how does sec^2x.tan^2x become (tan^3x/3)???? thx
It's a "recognition integral". You can differentiate (tan^3x)/3 to get sec^2x.tan^2x, or if you're not happy with that, use integration by parts:

Let I(x) = INT(sec^2x.tan^2x)

INT(sec^2x) = tanx and d(tan^2x)/dx = 2tanx.sec^2x

By parts,
I(x) = tanx.tan^2x - INT(tanx.2tanx.sec^2x)
I(x) = tan^3x - 2INT(sec^2x.tan^2x)
I(x) = tan^3x - 2I(x)
3I(x) = tan^3x
=> I(x) = (tan^3x)/3

as required
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14 years ago
#17
thx squishy
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14 years ago
#18
could someone please post a list of trig "recognition integrals" cos I never really spot them. Thank you x
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14 years ago
#19
There isn't really a definitive list, but there are a few identities and relationships you should always remember and try.

INT[cos^(n)x.sinx] = (-cos^(n+1)x)/(n+1)

INT[sin^(n)x.cosx] = (sin^(n+1)x)/(n+1)

INT[tan^(n)x.sec^2x] = (tan^(n+1)x)/(n+1)

INT[cot^(n)x.cosec^2x] = (-cot^(n+1)x)/(n+1)

sinx.cosx = ½sin2x, so INT[sinx.cosx] = -¼cos2x

These can be checked by differentiating.

Very useful identities:

1 + tan^2x = sec^2x

1 + cot^2x = cosec^2x

cos^2x = (cos2x + 1)/2
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