Question

Integrate

Answer 1

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

Answer 2

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

Answer 3

both equal 2

Answer 4

No, where did that identity come from?

Answer 5

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

Answer 6

yes

Answer 7

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.

Answer 8

There a 2 on the RHS.

Answer 9

What does RHS mean?

Answer 10

I meant that 1= cos x =2 cos^x/2. I forgot the 2 in the earlier post.
RHS means right hand side lol.

Answer 11

o

Answer 12

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

Answer 13

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

Answer 14

I got an answer of 2 for the cosine one?

Answer 15

Correct.

Answer 16

Where did the sin identity come from?

Answer 17

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.

Answer 18

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.

Answer 19

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?

Answer 20

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

Answer 21

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

Answer 22

Thanks.

Answer 23

no prob.