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Intergration Question

Hi

Could you show your working..or at least which method you would use to do the following:

∫ sin^3 x dx

sorry the symbol isn't very clear....it's "sine cubed x"

Thanks in advance !!

Reply 1

m:)ckel

14

sin^3 x = (sin^2 x)(sinx) = (1 - cos^2 x)(sinx) = sinx - (cos^2 x)(sinx)

So, ∫ sin^3 x dx = ∫ sinx dx - ∫ (cosx)^2(sinx)dx

The second integral is almost in the form of ∫ [f(x)^n].f '(x) dx = (1/n+1).[f(x)]^(n+1)
So, we can change it slightly to, .... + ∫ (cosx)^2 (-sinx) dx

So, the solution is:

-cosx + (1/3)cos^3 x + c

Reply 2

username9816

14

xskater

Hi

Could you show your working..or at least which method you would use to do the following:

∫ sin^3 x dx

sorry the symbol isn't very clear....it's "sine cubed x"

Thanks in advance !!

Evaluate: ∫ sin^3 x dx

Let u = cosx ---> du/dx = -sinx
---> dx = -du/sinx

∫ sin^3 x dx
= ∫ sin^2x.sinx
= ∫ (1 - cos^2x).sinx dx
= ∫ (1 - cos^2x).sinx. -du/sinx
= ∫ -(1 - u^2) du.
= ∫ u^2 - 1 du.
= u^3/3 - u + C
= cos^3x/3 - cosx + C
= cosx (cos^2x/3 - 1) + C

Reply 3

xskater

OP

6

cheers....the method works great BUT.....

i've never seen that rule before:
∫ [f(x)^n].f '(x) dx = (1/n+1).[f(x)]^(n+1)

Thanks again

Reply 4

m:)ckel

14

xskater

cheers....the method works great BUT.....

i've never seen that rule before:
∫ [f(x)^n].f '(x) dx = (1/n+1).[f(x)]^(n+1)

Thanks again

Have you got the Heinemann book P3 (or are you not doing Edexcel)?
Because, it's one of the standard forms for integrals.

Reply 5

xskater

OP

6

Cheers nima....

and i'm on AQA syllabus that's why... :frown:

Reply 6

m:)ckel

14

xskater

Cheers nima....

and i'm on AQA syllabus that's why... :frown:

Oh right....that's why- maybe you'll learn it elsewhere?

Reply 7

El Stevo

11

it isn't covered until at least p5...

Reply 8

BCHL85

12

mockel

Oh right....that's why- maybe you'll learn it elsewhere?

Oh no
You just use the parametric
let t = f(x), dt = f'(x)dx.
So u got the integral of (t^n).dt ---> it's easy to prove.
Cheers

Reply 9

m:)ckel

14

El Stevo

it isn't covered until at least p5...

Wow, why?
I can't see any point in that- it's not anything 'special'

Reply 10

m:)ckel

14

BCHL85

Oh no
You just use the parametric
let t = f(x), dt = f'(x)dx.
So u got the integral of (t^n).dt ---> it's easy to prove.
Cheers

Wait, what just happened here? :confused:

Reply 11

username9816

14

xskater

Cheers nima....

and i'm on AQA syllabus that's why... :frown:

Ditto.

That formula isn't officially taught on OCR either (I do AQA A-Level Maths, OCR AS Further Maths), but my teacher showed me it 2 days ago and says its very useful.

e.g.) Things like Integral: sin^3xcosx dx.

Can be evaluated much more easily using the formula. (Answer is sin^4x/4 + C).

Reply 12

Cirsium

19

xskater

Cheers nima....

and i'm on AQA syllabus that's why... :frown:

It's in the AQA P4 book - about chapter 8 if my memory serves

Reply 13

m:)ckel

14

Nima

Ditto.

That formula isn't officially taught on OCR either (I do AQA A-Level Maths, OCR AS Further Maths), but my teacher showed me it 2 days ago and says its very useful.

e.g.) Things like Integral: sin^3xcosx dx.

Can be evaluated much more easily using the formula. (Answer is sin^4x/4 + C).

Okay, so do you learn:

∫ [f '(x) / f(x)] dx = ln|f(x)| + c

???

Reply 14

El Stevo

11

mockel

Wow, why?
I can't see any point in that- it's not anything 'special'

i've finished p1-4 and haven't met it yet on p5... so either we don't formally cover it or it just hasn't cropped up yet... i saw the relationship in p2 when i was integrating by substitution (sin^2)xcosx, asked my teacher about it who said it was always true and used it as a shortcut ever since...

edit: its in a texbook? we don't get taught out of textbooks, our college makes its own notes for us to use....

Reply 15

elpaw

15

xskater

cheers....the method works great BUT.....

i've never seen that rule before:
∫ [f(x)^n].f '(x) dx = (1/n+1).[f(x)]^(n+1)

Thanks again

isnt it obvious? i.e. by inspection?

Reply 16

BCHL85

12

mockel

Wait, what just happened here? :confused:

I just think u don't need to learn it, just know how to do integral. It's just a way to do integral using parametric (as other methods like integral by part or by definition..)

Reply 17

m:)ckel

14

BCHL85

I just think u don't need to learn it, just know how to do integral. It's just a way to do integral using parametric (as other methods like integral by part or by definition..)