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# P3 Integration watch

1. I have to integrate the following with respect to x:

cos2xsinx

I found that when I used the substitution t = cosx i could then integrate the expression to obtain the correct answer but i was wondering if there was an easier way of solving this as usually the substition is given, but this time it was not.

Thanx
2. (Original post by Nylex)
cos 2x = 2cos^2 x - 1

Int cos 2xsin x dx = [(2cos^2 x - 1)sin x] dx

= (2sin xcos^2 x - sin x)dx

For the first bit, use u = cos x, du = -sin x dx (take the 2 and the - outside the integral).
GRRRRRRRRR YOU BEAT ME TO IT :P

and I wanted 2 practise my P3 integration so I don't go rusty!
3. (Original post by Hoofbeat)
GRRRRRRRRR YOU BEAT ME TO IT :P

and I wanted 2 practise my P3 integration so I don't go rusty!
Sorry. I've just deleted that post just now .
4. Why did you delete it Nylex?!
5. The way I'd do it is:

I(x) = INT[cos2x.sinx] dx

Using integration by parts,

I(x) = ½sin2x.sinx - INT[½sin2x.cosx] dx

Using parts again

I(x) = ½sin2x.sinx - (-¼cos2x.cosx - INT[-¼cos2x.(-sinx)] dx)
I(x) = ½sin2x.sinx - (-¼cos2x.cosx - ¼I(x))
I(x) = ½sin2x.sinx + ¼cos2x.cosx + ¼I(x)
¾I(x) = ½sin2x.sinx + ¼cos2x.cosx

And then you can get I(x)...maybe there's some brilliant shortcut, but I can't see it right now.
6. (Original post by Hoofbeat)
Why did you delete it Nylex?!
Cos he said he used the substitution t = cos x and that's what I did. He wanted a different way of doing it.
7. Using Integration by parts you get:

{ means integration sign

{cos2xsinx = -cosxcos2x - 2{sin2xcosx

Now

{sin2xcos = sin2xsinx - 2{cos2xsinx

Therefore,

{cos2xsinx = -cosxcos2x - 2[sin2xsinx - 2{cos2xsinx]
{cos2xsinx = -cosxcos2x - 2sin2xsinx + 4{cos2xsinx
{cos2xsinx - 4{cos2xsinx = -cosxcos2x - 2sin2xsinx
-3{cos2xsinx = -cosxcos2x - 2sin2xsinx
{cos2xsinx = (-1/3)[-cosxcos2x - 2sin2xsinx]

{cos2xsinx = (1/3)[2sin2xsinx + cos2xcosx]

I thinks thats right...
8. (Original post by Nylex)
Cos he said he used the substitution t = cos x and that's what I did. He wanted a different way of doing it.
oh rite. well I don't bother using the substitution (I know technically it is using the substitution) but I jsut recognise it's (=2sin xcos^2 x ) in the form f(x)^n.f'(x)
9. (Original post by integral_neo)
Using Integration by parts you get:

{ means integration sign

{cos2xsinx = -cosxcos2x + (1/2){sin2xcosx
Wait, shouldn't that be -cosx.cos2x - INT[2cosx.sin2x] dx?
10. (Original post by Squishy)
The way I'd do it is:

I(x) = INT[cos2x.sinx] dx

Using integration by parts,

I(x) = ½sin2x.sinx - INT[½sin2x.cosx] dx

Using parts again

I(x) = ½sin2x.sinx - (-¼cos2x.cosx - INT[-¼cos2x.(-sinx)] dx)
I(x) = ½sin2x.sinx - (-¼cos2x.cosx - ¼I(x))
I(x) = ½sin2x.sinx + ¼cos2x.cosx + ¼I(x)
¾I(x) = ½sin2x.sinx + ¼cos2x.cosx

And then you can get I(x)...maybe there's some brilliant shortcut, but I can't see it right now.
I dont think this is right...
11. (Original post by Hoofbeat)
oh rite. well I don't bother using the substitution (I know technically it is using the substitution) but I jsut recognise it's (=2sin xcos^2 x ) in the form f(x)^n.f'(x)
I don't think I was ever taught recognising f(x)^n.f'(x) and dealing with it like that.
12. (Original post by integral_neo)
I dont think this is right...
It is...I just checked.

¾I(x) = ½sin2x.sinx + ¼cos2x.cosx

Differentiate both sides:

cos2x.sinx + ½sin2x.cosx - ½sin2x.cosx - ¼cos2x.sinx
= ¾cos2x.sinx = ¾I'(x), as required

since cos2x.sinx = I'(x)
13. (Original post by Nylex)
I don't think I was ever taught recognising f(x)^n.f'(x) and dealing with it like that.
Me either, theres no point memorising rules like that really, you can just do it in your head. You KNOW that the differentials of sinx and cosx are related so you can just fiddle about
14. (Original post by imasillynarb)
Me either, theres no point memorising rules like that really, you can just do it in your head. You KNOW that the differentials of sinx and cosx are related so you can just fiddle about
Works for me .
15. (Original post by Hoofbeat)
GRRRRRRRRR YOU BEAT ME TO IT :P

and I wanted 2 practise my P3 integration so I don't go rusty!
You want some practice!! How would you go about integrating this:

sin 2xcos x

I used substitution but this method seems very long winded!
16. (Original post by the_anomaly)
You want some practice!! How would you go about integrating this:

sin 2xcos x

I used substitution but this method seems very long winded!
I'd use the wraparound method. Gets it done in a few lines.
17. (Original post by Squishy)
I'd use the wraparound method. Gets it done in a few lines.
What's the wraparound method??
18. (Original post by the_anomaly)
What's the wraparound method??
Look at my posts above...if you apply it, you get the integral as

-(sinx.sin2x + 2cosx.cos2x)/3
19. (Original post by the_anomaly)
You want some practice!! How would you go about integrating this:

∫sin 2xcos x

I used substitution but this method seems very long winded!
∫sin(2x)cosx
∫2sinxcosxcosx
∫2sinxcos²x
-(2/3)cos³x
=========
20. Integrate sec(x/2)tan(x/2).
I got the answer 2sec(x/2), I think its wrong. anyone who can trace whats wrong?

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