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# C3 Further Integration- Where did I go wrong? Watch

1. Q10) Where did I go wrong in my working out? The answer at the back of my textbook says it is 1/2Pi - 1.

Edit: Right, I took Penguin's, Keshim's and ghostwalker's advice and did v2 - v1. And this is what I achieved:

Could the textbook possibly be wrong?

EDIT: CASE CLOSED! =D
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2. (Original post by APonderingMind)
Q10) Where did I go wrong in my working out? The answer at the back of my textbook says it is 1/2Pi - 1.

tan^2(x) is the the lower curve, so your meant to do the integral of 1/2sec^2(x)-tan^2(x)
3. Well the way I'd do it is different to you..
I'd intergrate the equation y=1/2sec^2x - tan^2x between the limits of 0 and Pi/4 and then times this answer by two to get the area in between the two curves..
I would do it now but I'm off out so if you haven't worked it out later then quote me on here and I'll get round to helping out a bit better
In the meantime try my way? But good luck
4. (Original post by kashim91)
tan^2(x) is the the lower curve, so your meant to do the integral of 1/2sec^2(x)-tan^2(x)
I would still get -1 - 1/2Pi if I did it that way, not the correct answer
5. i can't remember how to do C3 maths...it was so long ago
6. The sec^2 curve is on the top in the region you are considering, so you want
7. (Original post by Penguinsaysquack)
Well the way I'd do it is different to you..
I'd intergrate the equation y=1/2sec^2x - tan^2x between the limits of 0 and Pi/4 and then times this answer by two to get the area in between the two curves..
I would do it now but I'm off out so if you haven't worked it out later then quote me on here and I'll get round to helping out a bit better
In the meantime try my way? But good luck
Hmm... still didn't get the same answer as the book.
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8. (Original post by APonderingMind)
Hmm... still didn't get the same answer as the book.
Your mistake is on the second line - the -1 should be a +1.
9. (Original post by APonderingMind)
Hmm... still didn't get the same answer as the book.
You made an error on your first line of working -tan^2(x) becomes 1-sec^2(x)
10. (Original post by ghostwalker)
You made an error on your first line of working -tan^2(x) becomes 1-sec^2(x)
(Original post by Mark13)
Your mistake is on the second line - the -1 should be a +1.

tan(x^2) = Sec(x^2) -1

so -(tanx^2) = -(secx^2 -1) = 1 - sec(x^2)

Oopsy... sign error. Thankies

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Updated: December 12, 2010
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