Question

integral of (sqrt)1+sin(theta)

Answer 1

\[\int\limits_{0}^{2\pi}\sqrt{1+\sin(\theta)}d \theta = -\cos/\sqrt{1-\sin(\theta)}\]

Answer 2

this is what they get

Answer 3

have i stumped everone haha

Answer 4

i believe this cannot be integrated.
at least as an indefinite integral

Answer 5

the answer above is wrong because its derivative does not equal sqrt(1+sin)

Answer 6

I have a very good thing about this one

\[I = \int_0^{2 \pi} \sqrt{1 +sin\theta}d \theta\]
\[I = \int_0^{2\pi} \sqrt{1 + sin{2\pi - \theta}}d\theta\]
\[I^2 = \int_0^{2\pi} \sqrt{1 - sin^2\theta}d\theta\]
\[I^2 = \int_0^{2\pi}cos\theta d\theta\]
\[I^2 = sin\theta|_0^{2\pi}\]
\[I = 0\]

Answer 7

ahh good answer
how did you get sin^2 ??
sin(2pi -theta) = -sin(theta)

Answer 8

right ... but wolfram doesn't agree with me 4sqrt2 is the wolfram answer

Answer 9

oh you squared I

Answer 10

yeah multiplied two equations maybe i have to find the integral I not I that means sqrtcos{theta}..

Answer 11

wolfram confuses me more ah it says complex answer to sqrtcosx form 0 to 2pi

Answer 12

yeah its 4sqrt(2), i just graphed it

Answer 13

wolfram one