# Integration

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#1
When using this integration rule, do I still divide by the derivative?
Last edited by Bigflakes; 3 months ago
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3 months ago
#2
(Original post by Bigflakes)
When using this integration rule, do I still divide by the derivative?
Not quite sure what you mean. The rule you have written (at the bottom) is correct. What do you mean by "divide by the derivative"?
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#3
(Original post by Pangol)
Not quite sure what you mean. The rule you have written (at the bottom) is correct. What do you mean by "divide by the derivative"?
Because when you are for example integrating sin2x you get -cos2x/2 because you are dividing by the derivative of the bracket (2x), so I’m asking if I have to divide ln(f(x)) by f’(x)?
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3 months ago
#4
(Original post by Bigflakes)
Because when you are for example integrating sin2x you get -cos2x/2 because you are dividing by the derivative of the bracket (2x), so I’m asking if I have to divide ln(f(x)) by f’(x)?
The best thing you could do would be to rewrite the integrand as (1/2)(2 cos(2x))/(1 + sin(2x)), and then factorise the 1/2 oustide the integral sign. The integrand is then of the exact form f'(x)/f(x).
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#5
(Original post by Pangol)
The best thing you could do would be to rewrite the integrand as (1/2)(2 cos(2x))/(1 + sin(2x)), and then factorise the 1/2 oustide the integral sign. The integrand is then of the exact form f'(x)/f(x).
Yes, that makes sense so I’m guessing that when using that rule, you don’t divide by the bracket. Thanks
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3 months ago
#6
(Original post by Bigflakes)
Yes, that makes sense so I’m guessing that when using that rule, you don’t divide by the bracket. Thanks
Still not quite sure what you mean (which bracket?), but it doesn't sound right! If you want to write up an example of what you were thinking of doing, we can try and be more helpful. (I should say that in general, your clear working and the fact that you show your attempts make it far easier to offer help that the average question poster.)
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3 months ago
#7
Although looking through this again now, if you mean "do I divide by 2?", then it is a yes. But I prefer to get an integrand of the exact form f'(x)/f(x) before integrating, it may take an extra line that doesn't do anything particularly clever, but it makes it very clear what you are doing.
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#8
(Original post by Pangol)
Although looking through this again now, if you mean "do I divide by 2?", then it is a yes. But I prefer to get an integrand of the exact form f'(x)/f(x) before integrating, it may take an extra line that doesn't do anything particularly clever, but it makes it very clear what you are doing.
Yes your way is better, thank you
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3 months ago
#9
(Original post by Bigflakes)
Yes your way is better, thank you
You need to be a bit careful when you're talking about "integration rules" - many people seem to invent their own rules for integration then get confused when things don't work out.

If you have something simple like f(x) = sin(kx) where k is a constant, then it's certainly true that f'(x) = kcos(kx), so if you are integrating cos(kx) then it would be correct to divide by k to get (1/k)sin(kx). But it's dangerous to assume that you can divide by arbitrary derivatives when you're integrating a ratio - it's almost always better to use a substitution until you're more confident
1
3 months ago
#10
you can use direct u-substitution
Last edited by Gaurang12; 3 months ago
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3 months ago
#11
(Original post by Gaurang12)
you can use direct u-substitution
You could. It’s quicker to spot that the numerator is just cos2x, and that 2sinxcosx is sin2x, so it’s in a form similar to f’(x) / f(x).
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3 months ago
#12
you are on the track of integration by Substitution
x=pi/4 x=pi/4
-- ---
| 1-2sin^2(x)/1+2sin(x)co(x)dx= | cos(2x)/1+sin(2) dx= u=sin(2x)-- apply integration by Substitution change variables to u=sin(2x) implies

----x=o ---x=0

implies du/dx=2cos(2x) remodify equation ---- 1

| 1/2du/dx / 1+u dx= 1/2(1+u)du integrate wrt u ----> 1/2In(1+u)= [ 1/2In(1+sin(2x)] x=pi/4 upper, lower=0
----

= 1/2In(2)
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