Question

evaluate the trig integral:
integral [0,pi/6] of sqrt(1+cos2x) dx

Answer 1

you know the formulas for cos 2x ?

Answer 2

I've found that it uses double angle formula - but I'm not familiar with it at all

Answer 3

the fact that it is equal to three things confuses me and i don't know which to pick

Answer 4

\(\Large \cos 2x = 2 \cos^2 x-1\)
use this.

Answer 5

Hmm...can you explain why you chose this one?

Answer 6

Is there a rule of thumb I guess is what i'm asking, if it's all just something you learn from experience then i guess i'm s.o.l. for now

Answer 7

to get rid of the 1

Answer 8

You could have chosen any of them and it would have worked out.

Answer 9

yes,
i considered all 3 formulas,
(remember we need the expression for "1+cos2x")

cos2x = cos^2x - sin^2 x
1+ cos2x = 1+cos^2 x - sin^2x
not much useful

cos 2x = 1-2sin^2 x
1+cos2x = 2-2sin^2x
hmm, can be used


cos 2x = 2cos^2x -1
1+cos2x = 2cos^2 x
very simple

so that sqrt (1+cos 2x) is just sqrt 2 cos x

Answer 10

If you use \(\cos(2x) = \cos^2(x)-\sin^2(x)\)Then you'd have: \[
1+\cos^2(x)-\sin^2(x) = \cos^2(x)+1-\sin^2(x) =\cos^2(x)+\cos^2(x) = 2\cos^2(x)
\]

Answer 11

beautiful, thanks hartnn and wio :)

Answer 12

I'll work on it from there