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anonymous

  • 4 years ago

Integrate

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{0}^{\pi/2}\sqrt{1+\cos t}dt\] \[\int\limits_{0}^{\pi/2}\sqrt{1+\sin t}dt\]

  2. anonymous
    • 4 years ago
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    Do you know that 1+cost= 2cos^2 (t/2) ?

  3. anonymous
    • 4 years ago
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    both equal 2

  4. anonymous
    • 4 years ago
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    No, where did that identity come from?

  5. anonymous
    • 4 years ago
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    Ok lets see do you know cos 2x= 2cos^2 x-1. ?

  6. anonymous
    • 4 years ago
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    yes

  7. anonymous
    • 4 years ago
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    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.

  8. anonymous
    • 4 years ago
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    There a 2 on the RHS.

  9. anonymous
    • 4 years ago
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    What does RHS mean?

  10. anonymous
    • 4 years ago
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    I meant that 1= cos x =2 cos^x/2. I forgot the 2 in the earlier post. RHS means right hand side lol.

  11. anonymous
    • 4 years ago
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    o

  12. anonymous
    • 4 years ago
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    Okay, I'll try to integrate it. So, the same goes for the sin one?

  13. anonymous
    • 4 years ago
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    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

  14. anonymous
    • 4 years ago
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    I got an answer of 2 for the cosine one?

  15. anonymous
    • 4 years ago
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    Correct.

  16. anonymous
    • 4 years ago
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    Where did the sin identity come from?

  17. anonymous
    • 4 years ago
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    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.

  18. anonymous
    • 4 years ago
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    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.

  19. anonymous
    • 4 years ago
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    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?

  20. anonymous
    • 4 years ago
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    Well its standard formula you just need to know it... :P

  21. anonymous
    • 4 years ago
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    It's not written anywhere in my textbooks. :(

  22. anonymous
    • 4 years ago
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    Thanks.

  23. anonymous
    • 4 years ago
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    no prob.

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