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# C4 trig/integration? Watch

1. Hi, I'm stuck on a few integration questions (which have trig in them) and would be grateful for some help!

1. Find the area between the x-axis and the curve y=xsin3x for 0<=x<=pi/3. Leave your answer in terms of pi. Find also the volume of the solid of revolution obtained by rotating this region about the x axis.

I was able to solve the first part through integration by parts, but I'm stuck with the second part. Here are my calculations so far:

(xsin3x)^2 = x^2 sin^2(3x)

So pi [integral sign] pi/3 and 0 (limits) x^2 sin^2(3x) dx

u = sin^2(3x)
du/dx= 6sin3xcos3x

dv/dx=x^2
v=1/3x^3

But then I get an even more complicated result?

2. Use the given substitution and then use integration by parts to complete the integration.

a) Integral of cos^-1(x) dx x=cos u

So dx/du = -sin u
and cos-1(X) = u

if I substitute that into the original equation:
integral of -usinu du

a = -u
da/du=-1

db/du= sinu
b=-cosu

so... ucosu - (integral of cosu )du
= ucosu - sinu + c
= cos^-1(x)cos(cos^-1(x)) - sin(cos^-1(x)) + c

...that's obviously wrong!

Thanks for your help
2. For 1 use the identity cos(2t)=1-2sin^2(t) then use IBP.

For the 2nd that last result may or may not be right (I haven't checked yet) but you can simplify it a lot. First off what is cos(cos^-1(x))? For sin(cos^-1(x)) draw a triangle and use pythagorus theorem.
3. (Original post by Bazinga?)
x
For the first question, you need to let u=x^2 then du/dx = 2x, hence your result becomes less complicated.

When doing integration by parts remember you're always trying to make it less complicated so always let the single x-variable be u (unless you're integrating ln(x) )

For the second question, let x=cos(u), then dx/du = -sin(u), from there, dx = -sin(u)du

Sub x=cos(u) to cos^-1(u) and you get cos^-1(cos(u)) which is equal to u (the coses cancel out)

because dx= -sin(u), sub dx into the original equation, you get -usin(u), which you can integrate by parts

I realise you did exactly the same as me in the last question, and I don't think it's wrong, just tidy it up a little and see what happens
4. Second one is right, just simplify it. As for the first one, switch your substitutions around, ie let u = x^2
5. (Original post by james22)
For 1 use the identity cos(2t)=1-2sin^2(t) then use IBP.

For the 2nd that last result may or may not be right (I haven't checked yet) but you can simplify it a lot. First off what is cos(cos^-1(x))? For sin(cos^-1(x)) draw a triangle and use pythagorus theorem.
Is that for the u = sin^2(3x) and du/dx= 6sin3xcos3x
Have I differentiated it wrong? As if I remember correctly, we were taught that y= sin^2(x) dy/dx= 2sinxcosx. So wouldn't this follow the same principle?

oh, would cos(cos^1(x)) just simplify to x?

but I'm not too sure how to simplify sin(cos^-1(x)) :/
6. (Original post by Anythingoo1)
For the first question, you need to let u=x^2 then du/dx = 2x, hence your result becomes less complicated.

When doing integration by parts remember you're always trying to make it less complicated so always let the single x-variable be u (unless you're integrating ln(x) )

For the second question, let x=cos(u), then dx/du = -sin(u), from there, dx = -sin(u)du

Sub x=cos(u) to cos^-1(u) and you get cos^-1(cos(u)) which is equal to u (the coses cancel out)

because dx= -sin(u), sub dx into the original equation, you get -usin(u), which you can integrate by parts

I realise you did exactly the same as me in the last question, and I don't think it's wrong, just tidy it up a little and see what happens
I think I'm a little bit lost with the cancelling out. I'm fine for cos^-1cos(x) = x. But not too sure about the sin(cos^-1(x)) ones.

(Original post by Ateo)
Second one is right, just simplify it. As for the first one, switch your substitutions around, ie let u = x^2
For the first question, I've switched the substitutions around...

u=x^2
du/dx=2x

dv/dx=sin^2(3x) or 1/2 - 1/2cos6x
v = 1/2x - 1/2xcos6x.

So...integral of x^2sin^2(3x) dx = something really complicated? And for the integral du/dx v, I have loads of x^2s.
7. (Original post by Bazinga?)
I think I'm a little bit lost with the cancelling out. I'm fine for cos^-1cos(x) = x. But not too sure about the sin(cos^-1(x)) ones.

For the first question, I've switched the substitutions around...

u=x^2
du/dx=2x

dv/dx=sin^2(3x) or 1/2 - 1/2cos6x
v = 1/2x - 1/2xcos6x.

So...integral of x^2sin^2(3x) dx = something really complicated? And for the integral du/dx v, I have loads of x^2s.
Forget about the limits for a second.

Repeat IBP for the second term.
8. (Original post by Bazinga?)
I think I'm a little bit lost with the cancelling out. I'm fine for cos^-1cos(x) = x. But not too sure about the sin(cos^-1(x)) ones.
For that I'd just leave it as it is, you wouldn't get deducted any marks for that in the exam (provided it's correct!)
9. (Original post by Bazinga?)
I'm not too sure how to simplify sin(cos^-1(x)) :/
10. (Original post by Ateo)
Forget about the limits for a second.

Repeat IBP for the second term.
er if you used IBP, would you get...

1/2x^3 - 1/2x^3cos6x - integral of (x^2 - x^2cos6x) dx?

So then integration again for x^2 - x^2cos6x?

so integral of x^2 = 1/3x^3

and IBP for the second part gives x^2/6 * sin6x - integral of x/3*sin6x dx.

So I have to do IBP again?!
11. (Original post by Anythingoo1)
For that I'd just leave it as it is, you wouldn't get deducted any marks for that in the exam (provided it's correct!)
And I probably wouldn't get something like this in the exam, but it would be nice to know how to do it.

(Original post by Ateo)
Sorry if it's blindingly obvious, but I still don't get it
12. (Original post by Bazinga?)
And I probably wouldn't get something like this in the exam, but it would be nice to know how to do it.

Sorry if it's blindingly obvious, but I still don't get it

I didn't bother with tex, so here you go, I know that I shouldn't really give you the full solution but it would be a bit too long to explain it bit by bit so here is only the indefinite integral. Its IBP twice.

13. (Original post by Bazinga?)
I think I'm a little bit lost with the cancelling out. I'm fine for cos^-1cos(x) = x. But not too sure about the sin(cos^-1(x)) ones.

For the first question, I've switched the substitutions around...

u=x^2
du/dx=2x

dv/dx=sin^2(3x) or 1/2 - 1/2cos6x
v = 1/2x - 1/2xcos6x.

So...integral of x^2sin^2(3x) dx = something really complicated? And for the integral du/dx v, I have loads of x^2s.
Have you got the answer for the first question? Had a go at it but not sure if I got the right answer

Actually don't worry someone posted the method above!

This was posted from The Student Room's iPhone/iPad App
14. (Original post by Ateo)

I didn't bother with tex, so here you go, I know that I shouldn't really give you the full solution but it would be a bit too long to explain it bit by bit so here is only the indefinite integral. Its IBP twice.

Thanks for your help. I'm going to try to do the problem again, before looking at your answer! I'll get back to you

(Original post by b4nanapancakes)
Have you got the answer for the first question? Had a go at it but not sure if I got the right answer

This was posted from The Student Room's iPhone/iPad App
It's 1/324 pi^2 (2pi^2-3)

Yeah, it's not a nice answer
15. (Original post by Bazinga?)
So I have to do IBP again?!
Yes
16. (Original post by Ateo)

I didn't bother with tex, so here you go, I know that I shouldn't really give you the full solution but it would be a bit too long to explain it bit by bit so here is only the indefinite integral. Its IBP twice.

Thanks for all your help, I tried it again and I spotted my mistake! When I was supposed to integrate, I differentiated
17. (Original post by Bazinga?)
Thanks for all your help, I tried it again and I spotted my mistake! When I was supposed to integrate, I differentiated
You're welcome. For any future questions you might do - in most cases, let u=x so that it cancels later on like in this example. You will, if you haven't already, encounter questions where you do IBP twice and are left with an integral that is the same as your original one (it would be one containing a trigonometric function). In that case you would take this integral over to the other side and divide by whatever number of the original integrals you have on the LHS.
18. (Original post by Ateo)
You're welcome. For any future questions you might do - in most cases, let u=x so that it cancels later on like in this example. You will, if you haven't already, encounter questions where you do IBP twice and are left with an integral that is the same as your original one (it would be one containing a trigonometric function). In that case you would take this integral over to the other side and divide by whatever number of the original integrals you have on the LHS.
So it's normal IBP only works one way round? i.e. if you have the substitutions the other way round, it may not work? As I encountered a few examples like this in the past and I always thought that I made a mistake or something.

Concerning your last point, could you give me an example of what you mean? As I don't quite follow what you are saying.
19. (Original post by Bazinga?)
So it's normal IBP only works one way round? i.e. if you have the substitutions the other way round, it may not work? As I encountered a few examples like this in the past and I always thought that I made a mistake or something.

Concerning your last point, could you give me an example of what you mean? As I don't quite follow what you are saying.
The whole point of IBP, is to reduce the integral into a simpler form, or if its cyclic to express the integral in terms of I
such as
If it becomes more complicated as you go, it won't work.
Do IBP twice on this to express it in terms of I, then rearrange.
20. (Original post by joostan)
The whole point of IBP, is to reduce the integral into a simpler form, or if its cyclic to express the integral in terms of I
such as
If it becomes more complicated as you go, it won't work.
Do IBP twice on this to express it in terms of I, then rearrange.
So you just leave it in terms of I?

Okay thanks!

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