anonymous
  • anonymous
Integrate
Mathematics
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anonymous
  • anonymous
Integrate
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits_{0}^{\pi/2}\sqrt{1+\cos t}dt\] \[\int\limits_{0}^{\pi/2}\sqrt{1+\sin t}dt\]
anonymous
  • anonymous
Do you know that 1+cost= 2cos^2 (t/2) ?
anonymous
  • anonymous
both equal 2

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anonymous
  • anonymous
No, where did that identity come from?
anonymous
  • anonymous
Ok lets see do you know cos 2x= 2cos^2 x-1. ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so take 1 to the other side you get 1+ cos 2x= 2cos^x. Replace 2x by x. You get 1+ cos x= cos^x/2.
anonymous
  • anonymous
There a 2 on the RHS.
anonymous
  • anonymous
What does RHS mean?
anonymous
  • anonymous
I meant that 1= cos x =2 cos^x/2. I forgot the 2 in the earlier post. RHS means right hand side lol.
anonymous
  • anonymous
o
anonymous
  • anonymous
Okay, I'll try to integrate it. So, the same goes for the sin one?
anonymous
  • anonymous
The two integrals are equal. And that is not an identity for sine.Only for cos. If you want to do the sin then 1+sinx= (sin (x/2) + cos(x/2))^2
anonymous
  • anonymous
I got an answer of 2 for the cosine one?
anonymous
  • anonymous
Correct.
anonymous
  • anonymous
Where did the sin identity come from?
anonymous
  • anonymous
Well expand the RHS and you get sin ^ x/2 + cos^2 x/2 + 2sin (x/2)cos(x/2) . sin ^2 x/2+ cos ^2 x/2=1. and 2sinx/2 cosx.2=sin x. So it becomes 1+sin x.
anonymous
  • anonymous
Do you know the property. \[\int\limits_{0}^{a}f(x)= \int\limits_{0}^{a}f(a-x)\]. If you apply this property then the second integral becomes the first one. So you don't need to do thhe second one it is equal to the first one.
anonymous
  • anonymous
Ah, I didn't know this property. I'm going to write it down. But how did you find the sin identity. I understand the reverse and it works but how did you think it up?
anonymous
  • anonymous
Well its standard formula you just need to know it... :P
anonymous
  • anonymous
It's not written anywhere in my textbooks. :(
anonymous
  • anonymous
Thanks.
anonymous
  • anonymous
no prob.

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