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STEP Integral watch

1. There is a question on the 1996 STEP II paper which looked possible (as they all do!) but it turns out, its not, at least for me anyway.

I have been able to do part a.

For the second part, I have got to

I = (from 0 to pi) ∫(sin t) / (√2(cos t/2) + √2(sin t/2) + 2) [dt.]

I am just confused on how to get it a. into a form which contains a pi/8 and b. into one term. No factorising I try, seems to work.
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cos(t/2) + sin(t/2) = √2 cos(t/2 - pi/4)
3. I = (from 0 to pi) ∫(sin t) / (2cos(t/2 - pi/4) + 2) [dt.]
= (from 0 to pi) ∫(sin t) / (2(cos(t/2 - pi/4) + 1)) [dt.]
= (from 0 to pi) ∫(sin t) / (2(2cos^2(t/4 - pi/8))) [dt.]
= (from 0 to pi) ∫(sin t) / (4cos^2(t/4 - pi/8)) [dt.]

Q.E.D.

Cheers!
4. (Original post by Vazzyb)
There is a question on the 1996 STEP II paper which looked possible (as they all do!) but it turns out, its not, at least for me anyway.

I have been able to do part a.

For the second part, I have got to

I = (from 0 to pi) ∫(sin t) / (√2(cos t/2) + √2(sin t/2) + 2) [dt.]

I am just confused on how to get it a. into a form which contains a pi/8 and b. into one term. No factorising I try, seems to work.
Since you know that:
√(1+cost) = √(1+cos²(t/2)-1) = √cos²(t/2) = cos(t/2)
√(1-cost) = √(1-(1-sin²(t/2)) = √sin²(t/2) = sin(t/2)

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Updated: December 19, 2005
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