Integration
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How do you integrate tan 2x? Tried to reduce it to the basic form but still don't know how to. Tried manipulating it with the formula of 1 + tan^2 2x = sec^2 2x, but alas, still can't do it.
Next i was given this question saying
Evaluate
(cos 2x - 1) /2 dx between range of 0 and 1. According to my teacher, whenever there's a word "evaluate" we must give the answer in surd form. However it seems to be impossible to leave this answer for this question in surd form. Even the answer provided by the teacher in the answer sheet "accidentally" gave the answer in decimal. Well, is there any way for me to convert it into surd form?
Thanks!
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int (tan 2x)
= int (sin 2x /cos 2x)
NOTE: remember d/dx(cos2x) = -2sin 2x
= (-1/2) int (-2sin 2x/ cos 2x)
= (-1/2)ln (cos 2x) + c
2nd issue... you heard wrongly.
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hi. thanks mikethm for providing your soln.
btw, you should put an absolute value around the cos2x, right after the ln. Thanks!
and yes, TS. For part 2 of your question, I think the limits are wrong. Cause it cant be a trig function with lim 0 -> 1, for it to be expressed in surds.
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2nd issue... you heard wrongly.
Yo.TS is right that the answer must be expressed in surd form if the calculator gives you decimal no. when questions explicitly asked you to evaluate.
Remember this vividly cuz the teachers in our schools were nagging at us for 1.5hours. It seems like all schools are emphazing on the "evaluate" questions.
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Originally posted by OHSheet:
hi. thanks mikethm for providing your soln.
btw, you should put an absolute value around the cos2x, right after the ln. Thanks!
and yes, TS. For part 2 of your question, I think the limits are wrong. Cause it cant be a trig function with lim 0 -> 1, for it to be expressed in surds.
in O level, int (1/x) is defined as ln(x) not ln|x|. This issue you have to raise with MOE, not me.
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Originally posted by bonkysleuth:
Yo.TS is right that the answer must be expressed in surd form if the calculator gives you decimal no. when questions explicitly asked you to evaluate.
Remember this vividly cuz the teachers in our schools were nagging at us for 1.5hours. It seems like all schools are emphazing on the "evaluate" questions.
surd form exist only when the limits are in special angles... eg. 0, pi/6, pi/4, pi/3 etc etc.
You can't get solutions in surds for 1 radian.
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For "O" level Add. Maths students who cannot remember the surd form answers for special angles of sin x, cos x and tan x
Example
sin 60 = 0.8660254038
what is the answer in surd form ?
Steps
1. Press "x^2" button and "=" button immediately on the calculator,
this gives an answer of 3/4
2. Next, square root (3/4) = [square root 3] /2 ie in surd form answer.
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Originally posted by Mikethm:
int (tan 2x)
= int (sin 2x /cos 2x)
NOTE: remember d/dx(cos2x) = -2sin 2x
= (-1/2) int (-2sin 2x/ cos 2x)
= (-1/2)ln (cos 2x) + c
2nd issue... you heard wrongly.
Still don't understand how integrating (-1/2) int (-2sin 2x/ cos 2x) gives you (-1/2)ln (cos 2x) + c.. what happens to tthe cos 2x at the bottom. Why is there suddenly a ln?
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Originally posted by anpanman:
Still don't understand how integrating (-1/2) int (-2sin 2x/ cos 2x) gives you (-1/2)ln (cos 2x) + c.. what happens to tthe cos 2x at the bottom. Why is there suddenly a ln?
For O level, integrating a fraction ( aka sin 2x/ cos 2x here).
There are 3 possible techniques.
1. Check if the top is similar to the differentiation of the bottom.
2. Check if Partial fraction can be applied to simplify the expression.
3. Check if the expression can be simplified to (ax+b)^n
For case (1), we are hoping to use the fact that...
d/dx[ln f(x)] = (f'x)/f(x)
Thus Int d/dx[ln f(x)] = Int (f'x)/f(x)
Therefore ln f(x) = Int (f'x)/f(x)
IN OTHER WORDS, IF THE TOP IS THE DERIVATIVE OF THE BOTTOM, THEN THE INTEGRAL WILL BE ln (BOTTOM)
So when we are trying to int (sin 2x/cos 2x)
We must check what is the differientation of the "cos 2x", side working will tell you that the derivative is "-2sin2x"
THUS WE NEED THE TOP TO LOOK LIKE "-2sin2x", this we manage by multiplying the expression by "-2/-2".
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Originally posted by Mikethm:
in O level, int (1/x) is defined as ln(x) not ln|x|. This issue you have to raise with MOE, not me.
Im referring it to the trig expression.
(-1/2)ln(cos2x) + c
ok. try substituting for x= pi/2. and try working it out.
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Originally posted by Mikethm:
For O level, integrating a fraction ( aka sin 2x/ cos 2x here).
There are 3 possible techniques.
1. Check if the top is similar to the differentiation of the bottom.
2. Check if Partial fraction can be applied to simplify the expression.
3. Check if the expression can be simplified to (ax+b)^n
For case (1), we are hoping to use the fact that...
d/dx[ln f(x)] = (f'x)/f(x)
Thus Int d/dx[ln f(x)] = Int (f'x)/f(x)
Therefore ln f(x) = Int (f'x)/f(x)
IN OTHER WORDS, IF THE TOP IS THE DERIVATIVE OF THE BOTTOM, THEN THE INTEGRAL WILL BE ln (BOTTOM)
So when we are trying to int (sin 2x/cos 2x)
We must check what is the differientation of the "cos 2x", side working will tell you that the derivative is "-2sin2x"
THUS WE NEED THE TOP TO LOOK LIKE "-2sin2x", this we manage by multiplying the expression by "-2/-2".
thanks mikethm!
TS, ill let you know on this. If im not wrong, when i was in sec4, my teacher taught me how to integrate functions like these with substitution..
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Originally posted by OHSheet:
Im referring it to the trig expression.
(-1/2)ln(cos2x) + c
ok. try substituting for x= pi/2. and try working it out.
Yup. I understood your point. However whether it is a trigo expression or not, the expression inside a logarithm function need to be positive. However my point is that the O level did not require the absolute value notation by telling you that int (1/u) is defined as "ln u" instead of the more specific "ln |u|" we do in university/poly calculus. My guess is that syllabus does not want to intro too much ideas and hence integrals which end up as "ln (something)" are not expected to have absolute value in them BUT the student is expected to know that (something)>0
Like my example... the syllabus say that int (1/x) = ln x + c.
Sub x = -2 and the expression is flawed. However students are expected to know that the intergral is true only when x>0 from their logarithm chapter. Hence the specific questioned asked by TS would require him to know that the intergral is true only for cos 2x>0.
If you disagree with this, you would be better served by raising your objection with MOE/SEAB/Cambridge as to why they structured their syllabus this way.
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Originally posted by anpanman:
How do you integrate tan 2x? Tried to reduce it to the basic form but still don't know how to. Tried manipulating it with the formula of 1 + tan^2 2x = sec^2 2x, but alas, still can't do it.
Next i was given this question saying
Evaluate
(cos 2x - 1) /2 dx between range of 0 and 1. According to my teacher, whenever there's a word "evaluate" we must give the answer in surd form. However it seems to be impossible to leave this answer for this question in surd form. Even the answer provided by the teacher in the answer sheet "accidentally" gave the answer in decimal. Well, is there any way for me to convert it into surd form?
Thanks!
The use of integration by substitution eg to integrate tan 2x is started by the teachers in the IP programme.
There are no actual past year "O" level Add Maths questions that asked students to solve integration by substitution. (Integration by substitution is tested in H2 Maths ie it is not tested in "O" level Add Maths. Though some school teachers eg Hua Yi and Swiss Cottage teahcers do teach the bright students the use of integration by substitution)
Instead, "O" level Add Maths students are required to integrate using the reversal method.
Example
(a) Differentiate x^3lnx
(b) Hence, integrate x^2 ln x dx
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Hi,
Thanks for your example. It is most typical of illustrating the relationship between differentiation and integration. It is sometimes set in H2 Maths, where (a) is about differentiating an expression and (b) is about integration by parts.
It is interesting to note that some sec sch teachers teach more advanced concepts to students.
On integration by substitution, my experience shows that even H2 Maths students struggle with this concept.
Thanks!
Cheers,
Wen Shih -
Example
(a) Differentiate x^3lnx
(b) Hence, integrate x^2 ln x dx
(a) Let y = x^3lnx
dy/dx = 3x^2lnx + x^3(1/x)
= 3x^2lnx + x^2
(b) Using the reversal method to integrate
Integrate x^2lnx dx
= Integrate [ (1/3) (3x^2lnx)]
= (1/3) Integrate 3x^2lnx
= (1/3) [ Integrate (3x^2lnx + x^2 - x^2)]
= (1/3) [ Integrate (3x^2lnx + x^2) - Integrate x^2 ]
Since the question uses the word, hence, we will need to make use of the
answer for part (a) ie Differentiate x^3lnx, we get 3x^2lnx + x^2, so when we
integrate 3x^2lnx + x^2, we get x^3lnx ie the reversal method
= (1/3) [ x^3lnx - x^3/3] + c
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Hi,
Thanks for your well-explained worked example!
If one integrates (b) directly, one is actually doing integration by parts :)
Cheers,
Wen Shih -
Hi Mr Wee,
Indeed, part (b) can be solved by integration by parts.
But "O" Level Add Maths students are not taught integration by parts.
They are only taught to use the reversal method to solve this type of question.
PS : Can I check with you why there is no H1 Maths TYS by Dyna this year ?
Thanks.
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Hi,
Yes, O-level maths does not cover integration by parts. I was making a remark to relate it to H2 maths, as a form of scaffolding to students who may be reading this thread :)
Erm, you may like to call Dyna to find out? In fact, I have helped them to classify last year's H1 exam questions at the end of 2008.
Thanks!
Cheers,
Wen Shih -
Hi Mr Wee,
Thank you for the prompt reply.
I have called Dyna just now and Jenny has confirmed that Dyna will not publish H1
Maths TYS this year.
So, the H1 Maths students will not have TYS this year.
Regards.