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# P3 Integration question - which method? watch

1. How would you solve this Integration question?
I used parts, but I had to do parts twice to it and then make it equal to I (which is some grade A work we did in class.)

However is there an easier way to do it.

The question is: The limits are pi/2 and 0.

{ 27sin2t.sint .dt =
2. (Original post by Nylex)
Well, since sin 2t.sin 2t = sin^2 2t, you can use the double angle formula for cos 2t: cos 2t = 1 - 2sin^2 t => sin^2 2t = 0.5(1 - cos 4t).

sorry I wrote it wrong, but Ive corrected it now.

It was actually 27sin 2t. sint
3. (Original post by BlueAngel)
How would you solve this Integration question?
I used parts, but I had to do parts twice to it and then make it equal to I (which is some grade A work we did in class.)

However is there an easier way to do it.

The question is: The limits are pi/2 and 0.

{ 27sin2t.sint .dt =
Use sin 2t = 2sin tcos t, so you have the integral of 27(2sin^2 tcos t) dt. Then you can use the substitution u = sin t, du = cos t dt.

INT 27(2sin^2 tcos t) dt

= 54 INT sin^2 tcos t dt

u = sin t, du = cos t dt:

= 54 INT u^2 du

= 54.(1/3)u^3

= 18sin^3 t

Putting in the limits:

= 18[(sin pi/2)^3 - (sin 0)^3]

= 18
4. (Original post by BlueAngel)
sorry I wrote it wrong, but Ive corrected it now.

It was actually 27sin 2t. sint
Yeah, no prob. Just seen that. See my post above.
5. (Original post by BlueAngel)
sorry I wrote it wrong, but Ive corrected it now.

It was actually 27sin 2t. sint

Cant you just use substitution straightaway, I mean

{27sin2t.sint.dt let U = sint so du/dt = cost and dt = 1/cost

27 { sin2t. U . 1/cost

27 { 2sintcost . U. 1/cost

27 { 2U.U = 27 { 2U^2 . du

Is that right????
6. (Original post by BlueAngel)
Cant you just use substitution straightaway, I mean

{27sin2t.sint.dt let U = sint so du/dt = cost and dt = 1/cost

27 { sin2t. U . 1/cost

27 { 2sintcost . U. 1/cost

27 { 2U.U = 27 { 2U^2 . du

Is that right????
I'm not following that.. how are you going from du/dt = cos t to dt = 1/cos t? Full worked answer above btw.
7. (Original post by Nylex)
I'm not following that.. how are you going from du/dt = cos t to dt = 1/cos t? Full worked answer above btw.

Yes, I just used this method and it worked I got 18.

If du/dt = cost you want to replace dt dont you.

So you bring the du to the other side, so dt = du/cost. Which also is 1/cost .du.

Get it?
8. you can actually spot that since sin t differentiates to cos t, so the integral of sin^2 t cos t will go exactly to some multiple of sin^3 t because of the chain rule. it's an exact integral.
9. (Original post by BlueAngel)
Yes, I just used this method and it worked I got 18.

If du/dt = cost you want to replace dt dont you.

So you bring the du to the other side, so dt = du/cost. Which also is 1/cost .du.

Get it?

This is how I did it, wow Its easier than the teachers method.

{27sin2t.sint.dt
let U = sint, so du/dt = cost, and dt = 1/cost

27 { sin2t. U . 1/cost

27 { 2sintcost . U. 1/cost

27 { 2U.U = 27 { 2U^2 . du

Integrate 2U^2 = 2U^3/3.27.

Replace U with sint. So you get 2sint^3/3

Put the limits into it, which are pi/2 and 0.

And you get 18 units squared.
10. (Original post by 4Ed)
you can actually spot that since sin t differentiates to cos t, so the integral of sin^2 t cos t will go exactly to some multiple of sin^3 t because of the chain rule. it's an exact integral.

Yeah youre right, thanx for pointing that out.
11. (Original post by 4Ed)
you can actually spot that since sin t differentiates to cos t, so the integral of sin^2 t cos t will go exactly to some multiple of sin^3 t because of the chain rule. it's an exact integral.
Oh crap, yeah. Inverse chain rule. Damn it.
12. PHEW! I dont know why you guys are all worried? You all seem to know everything about everything!!

Im just gonna try lol .. off to do sum review questions
13. How about using the product-to-sum identity?

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