You are Here: Home >< Maths

# p3 help, rep will be given watch

1. if you have a p3 book its ex 4e q 47

use the sub x=3sint to show that

intergration limits (3,0) x^2(9 - x^2)^0.5 dx = 81(pie)/16

using theories below and including p3 level please
2. I remember this one! Basically, you sub x=3sint into the equation, and then mess it around a bit and it will make something very nice to integrate. I would do it for you but I've had loads of codeine
3. x = 3sint
dx = 3cost dt

Integral becomes:
9sin²t.(9-9sin²t).(3cost) dt
9sin²t.3cost.3cost dt [using 1-sin²t=cos²t]
81sin²tcos²t dt
20.25(4sin²tcos²t) dt
20.25(sin2t)^2 dt [using sin2t=2sintcost]
20.25(1-cos4t)/2 dt [using cos4t=1-2(sin2t)^2]
10.125 [t - (sin4t)/4]
Limits become:
3 -> pi/2
0 -> 0
So integral becomes:
10.125[(pi/2 - 0) - (0)] = 10.125pi/2 = 81pi/16

Hmm.. that was long-winded. I think there's a better way to do it, but I'm just not seeing it right now.
4. thanks dvs but anyone got a better method
5. (Original post by BLUREMI)
Use the sub. x = 3sint to show that:
Int. [limits (3,0)] x^2(9 - x^2)^(1/2) dx = (81pi)/16
Let x = 3sint --> dx/dt = 3cost --> dx = 3cost dt

Hence: Int. x^2(9 - x^2)^(1/2) dx
= Int. 9sin^2t(9 - 9sin^2t)^(1/2) . 3cost dt
= Int. 9sin^2t[9(1 - sin^2t)]^(1/2) . 3cost dt
= Int. 9sin^2t(9cos^2t)^(1/2) . 3cost dt
= 81 Int. sin^2tcos^2t dt
= (81/4) Int. 4sin^2tcos^2t dt
= (81/4) Int. [2sintcost]^2 dt
= (81/4) Int. sin^2(2t) dt
= (81/4) Int. 1/2(1 - cos4t) dt
= (81/8) [t - (sin4t)/4] + k
= 81t/8 - 81(sin4t)/32 + k
= [324t - 81(sin4t)]/32 + k

To find the new limits:

When x = 3: --> 3 = 3sint --> sint = 1 --> t = Pi/2
When x = 0: 0 = 3sint --> sint = 0 --> t = 0

Hence: Int. [limits (3,0)] x^2(9 - x^2)^(1/2) dx
= {324t - 81(sin4t)]/32} [limits Pi/2 and 0]
= (162Pi)/32
= (81Pi)/16

Nima
6. That's the same way I did it. So I take it there's no simpler way?
7. (Original post by Nima)
Sorry lol, I didn't copy your post, honest. I saw the thread starter's post and tried to do it in the simplest way I could. I think it is the simplest way - Originally I was going to do it by parts until I spotted that 4sin^2tcos^2t is useful...

Nima
lol, I saw you many times did like that. I thought you just try to do whatever you see, don't you?
8. (Original post by Nima)
Sorry lol, I didn't copy your post, honest.
I didn't mean it like that. I'm sorry if it sounded as if I did.
9. do u think questions could be that hard in the p3 exam?

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: February 17, 2005

### University open days

Wed, 21 Nov '18
• Buckinghamshire New University
Wed, 21 Nov '18
• Heriot-Watt University
Wed, 21 Nov '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams