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Differentiation & Integration: A Hence Question

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Tuition given in the topic of A-Maths Tuition Questions from the tuition centre desk of Miss Loi at 6:35 pm (Singapore time)

Ok holiday’s over, had enough fun with the different standards of foreign talents, time to get back to your studies!

As a maths-related blog, many of you here have been deprived of a question containing that most iconic of mathematical symbols: the sexy and alluring integral symbol with her … errm … 10-2-10 curvaceous figure.

Hence, Miss Loi shall not deprive you further …

Differentiate ln(cos 2x) with respect to x. Hence, or otherwise, evaluate

int{0}{pi/6}{tan 2x} dx

Please, please note the word Hence in the question! The words “or otherwise” are usually placed there just for show, unless you like to do things the long way. Hence Miss Loi terms these Hence Questions.

If that’s not clear enough, the Hence is meant to remind you to use information from your answer/workings in an earlier part to solve the next part. Sounds easy but sometimes it’s not so straightforward and you’ll usually need to think of one more step before attaining enlightenment.

Differentiation and Integration are a match made in heaven. So unless they are having a tiff, there’s a good chance you’ll see them together in the same question. Miss Loi has observed that many students got so enchanted by the sexy integral sign that they totally forgot about their differentiation techniques, and vice versa for those with more … errmmm … ‘differential’ tastes.

So whenever you spot the word Hence, remember to 看前面 (watch your front again)!!!

頑張って!!!

Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

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Comments & Reactions

2 Comments

  1. fggffggf's Avatar
    fggffggf commented in tuition class


    2008
    Jun
    15
    Sun
    12:09am
     
    1

    {d/dx}{ln(cos 2x)}= (1/{cos 2x}) (- sin 2x) (2)
    {}= {-2sin 2x}/{cos 2x}=-2tan 2x

    Yup remember your {d/dx}(ln u) = {1/u}{du/dx} here!

    .'.tan 2x={-1/2}{d/dx}{ln(cos 2x)}={d/dx}({-1/2}{ln(cos 2x)})

    This is the key part of this hence question: expressing tan 2x as a differential using your earlier answer. May Heaven help you if you're trying to integrate tan 2x on your own! (At least if you're an O-Level student anyway :P)

    .'.int{0}{pi/6}{tan 2x} dx = delim{[}{{-1/2}{ln(cos 2x)}}{]}^{pi/6}_0

    ={-1/2}{ln(cos (2*{pi/6}))}+1/2{ln(cos (2*0))}

    =-1/2{ln(1/2)}+1/2{ln(1)}

    =-1/2{ln(1/2)}=0.34657 (corr. to 5 sig. fig.) CORRECTO!

    A little thought on the integral sign: it looks more like the figure just under the strings of a violin, but I do agree that it is rather like a tangent curve. My E+A-math teacher says tangent curves are like "pretty ladies in a beauty pageant".

    OMG the pageant must be held outdoors since tangent curves tend to reach high into infinity! Sorry lame math jokes I know ...

    And when I put E and A together I instantly thought of the slogan of EA games (I don't play any EA games, though): EA Games - Challenge Everything.

    Waiting for more math challenges! Stay tuned 😉

    p.s. the math expressions are all stuck together because when I tried to separate them there was always some problems with the display.

  2. sham sihabdeen's Avatar
    sham sihabdeen commented in tuition class


    2008
    Jul
    22
    Tue
    3:49pm

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