anonymous
  • anonymous
Integrate
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.