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# Integrating Cos(X)Sin(X) three ways. watch

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1. Well, as you can tell, its another test i've got today, so my maths ability has dissapeared.
Please can someone remind me how to integrate this using three methods?
2. what do you mean by three methods?

a) by parts
b) using the identity sin(x)cos(x)=sin(2x)/2
c) changing cos(x) = √(1-sin²(x)) and using a substitution u = sin(x)

???????????????????????????????? ??????????
3. (Original post by elpaw)
what do you mean by three methods?

a) by parts
b) using the identity sin(x)cos(x)=sin(2x)/2
c) changing cos(x) = √(1-sin²(x)) and using a substitution u = sin(x)

???????????????????????????????? ??????????
Thanks.
Thats exactly what i meant.
My brain switches off before an exam, so i couldn't remember.
And i did them last thing yesterday!
4. hmm i need to get back up there with the maths - it's all slipped away from me this year... i couldn't even remember one method of doing that!
5. Just thought i'd give these a go.....

I = ∫sin(x)cos(x)dx

(a) by parts

u = cos(x) v' = sin(x)
u' = -sin(x) v = -cos(x)
=> I = -cos²(x) -∫sin(x)cos(x)dx = -cos²(x) - I
=> 2I = -cos²(x)
=> I = -cos²(x)/2 +C == sin²(x)/2 +C'

(b) using the identity sin(x)cos(x)=sin(2x)/2

I = ∫sin(2x)/2 dx
= -1/4 cos(2x) +C
(you can go on):
I = -1/4 (cos²(x) - sin²(x)) + C
= -1/4 (2cos²(x) - 1) + C
= -cos²(x)/2 + C' == sin²(x)/2 + C''

hence (b) == (a)

(c) changing cos(x) = √(1-sin²(x)) and using a substitution u = sin(x)

I = ∫√(1-sin²(x))sin(x)dx

u = sin(x) => du = cos(x) dx = √(1-u²) dx
=> dx = du/√(1-u²)

I = ∫√(1-u²)u/√(1-u²) du
= ∫u du
= u²/2 + C
= sin²(x)/2 +C == -cos²(x)/2 +C'

hence (c) == (b) == (a)
6. surely the easiest way is let u=sinx, then the integral becomes u.du which is 1/2sin^2x + C
7. (Original post by Ralfskini)
surely the easiest way is let u=sinx, then the integral becomes u.du which is 1/2sin^2x + C
The easiest way is to do it by inspection, recognising that it's sort of in the form f'(x)f(x). You can just integrate as you would a normal function, checking that you get the negatives correct.

[Integral]sin(x)cos(x) = (1/2)sin^2(x) + C

OR

[Integral]sin(x)cos(x) = -(1/2)cos^2(x) + D

Ben
8. jus use the substitution u = sinx

then du = cosx dx

∫u du

u^2 * (1/2)

(sin^2 x )/2

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