## anonymous 4 years ago Integrate

1. anonymous

$\int\limits_{0}^{\pi/2}\sqrt{1+\cos t}dt$ $\int\limits_{0}^{\pi/2}\sqrt{1+\sin t}dt$

2. anonymous

Do you know that 1+cost= 2cos^2 (t/2) ?

3. anonymous

both equal 2

4. anonymous

No, where did that identity come from?

5. anonymous

Ok lets see do you know cos 2x= 2cos^2 x-1. ?

6. anonymous

yes

7. 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.

8. anonymous

There a 2 on the RHS.

9. anonymous

What does RHS mean?

10. 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.

11. anonymous

o

12. anonymous

Okay, I'll try to integrate it. So, the same goes for the sin one?

13. 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

14. anonymous

I got an answer of 2 for the cosine one?

15. anonymous

Correct.

16. anonymous

Where did the sin identity come from?

17. 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.

18. 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.

19. 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?

20. anonymous

Well its standard formula you just need to know it... :P

21. anonymous

It's not written anywhere in my textbooks. :(

22. anonymous

Thanks.

23. anonymous

no prob.