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tammypotato

plz can u integrate sec^4 xdx thanks!

Yeah plzzzzzzzzz I've been stuck on it for ages

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

It'sPhil...

sec^4x = sec^2x(1 + tan^2x)

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

=> INT = tanx + (tan^3x)/3 + C

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

=> INT = tanx + (tan^3x)/3 + C

lol thank u soooo much

It'sPhil...

sec^4x = sec^2x(1 + tan^2x)

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

=> INT = tanx + (tan^3x)/3 + C

= 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?

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

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

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.

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.

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?

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

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

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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