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Basic Integration Help (Core 4)

Hi guys, really struggling with these simple questions although I can do the 'harder' questions.



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Reply 1
These questions don't require any real tricks, you just need to know your derivatives of trig functions. Some of them will probably be in your formula book in fact. If you know the derivatives of tan x, sec x, cosec x and cot x then you'll be able to do these questions. The only snag is the fact that you have 5x or 7x instead of just x, but you should know how to get past that.
Reply 2
What is a list of the integrals I need to revise? I mean if the equation wants the integral of sec5xtan5x and I know the integral of sec5x and the integral of tan5x how do I apply this to the original question? Thanks a lot mate
Reply 3
Original post by Mousebudden
What is a list of the integrals I need to revise? I mean if the equation wants the integral of sec5xtan5x and I know the integral of sec5x and the integral of tan5x how do I apply this to the original question? Thanks a lot mate


These are from the Edexcel Formula Booklet
First one put them in terms of sin and cos than use substitution of u= cosx this will cancel out sin x and than the answer should be obvious. Same with the second one except it's now u = sinx. For last one its the cos2x double angle formula
Original post by nuodai

Original post by nuodai
These questions don't require any real tricks, you just need to know your derivatives of trig functions. Some of them will probably be in your formula book in fact. If you know the derivatives of tan x, sec x, cosec x and cot x then you'll be able to do these questions. The only snag is the fact that you have 5x or 7x instead of just x, but you should know how to get past that.


Are you sure? because I am pretty sure what you are saying is wrong
(edited 13 years ago)
Just realised you could also use trig identities
Integral of sec(5x)tan(5x)
1. Set u =5x
2. Thus du = 5dx
3. Substitute:
Integral of (1/5)sec(u)tan(u)du
4. Set v = sec(u)
5. Thus dv = sec(u)tan(u)
6. Integral of (1/5)(1dv)
7. Integral of (1/5)(1dv) = (1/5)v
8. Substitute v back in.
=(1/5)sec(u) + C
9. Substitute u back in.
=(1/5)sec(5x) + C
Integral of csc(7x)cot(7x)
1. Set u = 5x
2. Thus du = 5dx
3. Substitute:
Integral of (1/7)csc(u)cot(u)
4. Set v = csc(u)
5. Thus dv = -csc(u)cot(u)
6. Integral of (-1/7)(1dv)
7. Integral of (-1/7)(1dv) = (-1/7)v
8. Substitute v back in.
=(-1/7)csc(u) + C
9. Substitute u back in.
= (-1/7)csc(7x) + C
Just realised they are actually in the formula booklet. Oh well it works as proof
Reply 10
Original post by loki276
Are you sure? because I am pretty sure what you are saying is wrong


Yes I'm sure. The derivative of sec5x\sec 5x is 5sec5xtan5x5\sec 5x \tan 5x, for example, and the other two integrands are also constant multiples of derivatives of simple trig functions.
Original post by nuodai

Original post by nuodai
Yes I'm sure. The derivative of sec5x\sec 5x is 5sec5xtan5x5\sec 5x \tan 5x, for example, and the other two integrands are also constant multiples of derivatives of simple trig functions.


Yeah just saw it
Integral of (csc(7x))^(2)
1. Set u = 7x
2. Thus du = 7dx
3. Substitute:
Integral of ((1/7)csc(u))^(2)
**Integral of (csc(u))^(2) = -cot(u)**
= (1/7)-cot(u)
4. Substitute u back in.
= -(1/7)cot(7x)
Reply 13
What I don't understand is how will the derivatives help me to find the integral? Sorry if it sounds really stupid.


Original post by CaliforniaKid
Integral of csc(7x)cot(7x)
1. Set u = 5x
2. Thus du = 5dx
3. Substitute:
Integral of (1/7)csc(u)cot(u)
4. Set v = csc(u)
5. Thus dv = -csc(u)cot(u)
6. Integral of (-1/7)(1dv)
7. Integral of (-1/7)(1dv) = (-1/7)v
8. Substitute v back in.
=(-1/7)csc(u) + C
9. Substitute u back in.
= (-1/7)csc(7x) + C


I'm with you up untill Step 3, when you take out 1/7. How are you able to do that?

Thanks a lot guys! All Repped
Keep in mind the method I used in the first two problem is the long way.

By now you should know:
the integral of secxtanx = secx
the integral of cscxcotx = -cscx

Original post by Mousebudden
What I don't understand is how will the derivatives help me to find the integral? Sorry if it sounds really stupid.

I'm with you up untill Step 3, when you take out 1/7. How are you able to do that?

Thanks a lot guys! All Repped


The integral is the opposite of the derivative.
Example:
Derivative of x^3
=3x^2
Integral of 3x^2
=x^3

I multiplied by (1/7) because there is a (7x). If there was a 7 in front of the problem I would not need to multiply by (1/7).
(edited 13 years ago)
Its just the reverse of it if that makes sense
Reply 16
The integrals of those are in your formula book. I mean obviously its useful to remember them but, they are the easiest integrations you can get asked.
Reply 17
As the derivative of sec(x) is sec(x)tan(x), the integral of sec(x)tan(x) is sec(x)?

Therefore, to refer to the original question, the integral of sec(5x)tan(5x) is sec(5x) ?
Original post by Mousebudden
As the derivative of sec(x) is sec(x)tan(x), the integral of sec(x)tan(x) is sec(x)?

Therefore, to refer to the original question, the integral of sec(5x)tan(5x) is sec(5x) ?

Not quite, you're out by a factor. Note that ddx[sec(5x)]=5sec(5x)tan(5x)\dfrac{d}{dx}[\sec (5x)] = 5\sec (5x)\tan (5x).
Reply 19
Original post by Farhan.Hanif93
Not quite, you're out by a factor. Note that ddx[sec(5x)]=5sec(5x)tan(5x)\dfrac{d}{dx}[\sec (5x)] = 5\sec (5x)\tan (5x).


Got it. Thanks mate.

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