Turn on thread page Beta
 You are Here: Home

P3 integration AGAIN!!! watch

1. Hi

I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

The integral of:

cosx/(5-3sinx)

Would it also be possible to tell me how you clever people do this question???

There will probably be more questions later - but not for the moment!!!

Thanks
2. (Original post by Silly Sally)
Hi

I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

The integral of:

cosx/(5-3sinx)

Would it also be possible to tell me how you clever people do this question???

There will probably be more questions later - but not for the moment!!!

Thanks
for this one use the substitution t=tan(x/2) (so get sinx and cosx in terms of t)
by the way,what syallabus is this(its about p5 for edexcel)
3. (Original post by Silly Sally)
Hi

I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

The integral of:

cosx/(5-3sinx)

Would it also be possible to tell me how you clever people do this question???

There will probably be more questions later - but not for the moment!!!

Thanks
Well if you differentiate ln(5-3sinx) you'd get -3cosx/5-3sinx.

So you want -1/3 cosx/5-3sinx + c.
4. (Original post by IntegralAnomaly)
for this one use the substitution t=tan(x/2) (so get sinx and cosx in terms of t)
by the way,what syallabus is this(its about p5 for edexcel)
I am doing Edexcel - that can't be right!!! About P5 - this is from a P3 revision book- if you do Edexcel maths you may recognise the book - "revise for Pure mathematics 3" by Heinemann p28. I mean if you say it is P5 - i believe you - it is just really strange...
5. (Original post by theone)
Well if you differentiate ln(5-3sinx) you'd get -3cosx/5-3sinx.

So you want -1/3 cosx/5-3sinx + c.
ohh crap,yes this method is the proper method,forget what i said(theone ur answer is not correct)
6. Do you do integration backwards - i.e. differentiate and see what you get and then find integral - cos i do that when it is simpler equation - but i can't do it with harder equations
7. (Original post by Silly Sally)
Do you do integration backwards - i.e. differentiate and see what you get and then find integral - cos i do that when it is simpler equation - but i can't do it with harder equations
use the substitution u=5-3sinx
8. (Original post by IntegralAnomaly)
use the substitution u=5-3sinx
Ahh - I understand substitution. Thanks for the tip.

Btw - i would like to ask you all 2 questions rearding integration:

1. When you look at a question, what is your process of thinking to deduce which method to use - cos i have difficulty looking at a question and finding a method that will click (this may sound quite a silly naf question but anything will help!!!)

2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

Hope that makes sense???
9. (Original post by Silly Sally)
Hi

I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

The integral of:

cosx/(5-3sinx)

Would it also be possible to tell me how you clever people do this question???

There will probably be more questions later - but not for the moment!!!

Thanks
U=5-3SinX therefore U'=-3CosX which means dx = -du/3CosX

Therefore CosX/(5-3SinX) -du/3CosX:

= -1/3 * Integral of u^-1
=-1/3lnU
=-1/3ln(5-3SinX) + C

Sorry, but I don't have the fancy shmancy keyboard signs.
10. (Original post by Silly Sally)
Ahh - I understand substitution. Thanks for the tip.

Btw - i would like to ask you all 2 questions rearding integration:

1. When you look at a question, what is your process of thinking to deduce which method to use - cos i have difficulty looking at a question and finding a method that will click (this may sound quite a silly naf question but anything will help!!!)

2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

Hope that makes sense???
For part 2. Use the chain rule e.g:

Y=(CosX)^4 say that U=CosX therefore Y=U^4

dy/du = 4U^3

You know that U=CosX therefore:

du/dx = -SinX

The chain rule says that dy/dx = dy/du * du/dx therefore:

dy/dx = 4U^3 * -SinX and replace the U with CosX leaving:

-4CosX^3SinX

For part 1, its just practice.
11. (Original post by Bhaal85)
For part 2. Use the chain rule e.g:

Y=(CosX)^4 say that U=CosX therefore Y=U^4

dy/du = 4U^3

You know that U=CosX therefore:

du/dx = -SinX

The chain rule says that dy/dx = dy/du * du/dx therefore:

dy/dx = 4U^3 * -SinX and replace the U with CosX leaving:

-4CosX^3SinX
.
For the part 2 you were answering, i was referring to integration rathe than differentiation.
12. (Original post by Silly Sally)
For the part 2 you were answering, i was referring to integration rathe than differentiation.
Oh **** , sorry, what can I say, late night.
13. (Original post by Silly Sally)
Hi

I really need help with P3 integration - i don't like it and i can't do it!!! So here is the question:

The integral of:

cosx/(5-3sinx)

Would it also be possible to tell me how you clever people do this question???

There will probably be more questions later - but not for the moment!!!

Thanks
u = 5-3sinx
du = -3cosx dx

integrate -3/u = -3lnu + c
=-3ln(5-3sinx) + c

is that right?
14. (Original post by Silly Sally)
Ahh - I understand substitution. Thanks for the tip.

Btw - i would like to ask you all 2 questions rearding integration:

1. When you look at a question, what is your process of thinking to deduce which method to use - cos i have difficulty looking at a question and finding a method that will click (this may sound quite a silly naf question but anything will help!!!)

2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

Hope that makes sense???
u have the recognise(unlike i did above) the different forms of integration u can have,if u have something like f'(x)/f(x) then the integral will be ln|f(x)|+c,
eg. ∫(sec²x/tanx)dx,u know that the derivative of tanx is sec²x,and this is of the form f'(x)/f(x) and so the answer is ln|tanx|+c.
If u still are not confident with spotting it u can always use a substitution (u=f(x)).
so have a go at this one:∫1/(xlnx)dx.
If u have two completely different functions of x (x²e^x) then u should recognise that u have to use integration by parts.
As for the trigonometric ones u have to try and get it in the form of f'(x)f(x).
And then u can use a substitution u=f(x).
15. (Original post by Bhaal85)
Oh **** , sorry, what can I say, late night.
Its ok - it confused me a bit more than usual - and i am easily confused!!!

So do you have a method for integration with high powers that i was referring to in a previous post? Would really appreciate it!!!
16. (Original post by imasillynarb)
u = 5-3sinx
du = -3cosx dx

integrate -3/u = -3lnu + c
=-3ln(5-3sinx) + c

is that right?
no.look again at the cosx term.
17. (Original post by Bhaal85)
U=5-3SinX therefore U'=-3CosX which means dx = -du/3CosX

Therefore CosX/(5-3SinX) -du/3CosX:

= -1/3 * Integral of u^-1
=-1/3lnU
=-1/3(5-3SinX) + C

Sorry, but I don't have the fancy shmancy keyboard signs.
Typo:

-1/3ln(5-3SinX) + C
18. (Original post by imasillynarb)
u = 5-3sinx
du = -3cosx dx

integrate -3/u = -3lnu + c
=-3ln(5-3sinx) + c

is that right?
Nearly - cos its integration - it is -1/3ln(5-3sinx)
19. (Original post by Silly Sally)
2. when you have a trig eqn raised to a power more than 2 and 3, what do you do to tackle the question. My teacher was saying that if the power is odd, e.g. (cosx)^5, then you leave a cosx outside and convert the (cosx)^4 to expression containing sinx so you can use substitution. So do you use a "special method" for even power functions or a mixture of funtions e.g. (sinx)^2(cosx)^3???

with an odd power, e.g. sin^n x, you split it into sinx sin^(n-1) x, which becomes sinx - sinx cos^(n-1) x, which is a straightforward integral to cos^n x -cos x

with an even power, you have to use the cos(2x) forumla

e.g. sin²x = (1-cos2x)/2
cos²x = (1+cos2x)/2

so, eg. sin^2n x = (sin²x)^n = ((1-cos2x)/2)^n, and expand binomially
20. (Original post by Silly Sally)
Its ok - it confused me a bit more than usual - and i am easily confused!!!

So do you have a method for integration with high powers that i was referring to in a previous post? Would really appreciate it!!!
I don't think that I have been taught how to do the integral of that trig:

If I were doing it the method I would do, I would get:

-CosX^6/6SinX + C which doesn't look right.

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 15, 2004
Today on TSR

Are exams rubbish?

Is the exam system out of date?

University open days

• University of Chichester
Thu, 25 Oct '18
• Norwich University of the Arts
Fri, 26 Oct '18
• University of Lincoln