Question

integral of rad(1+sin2x+(cos2x/rad(1+sin2x))^2) 0 12 years ago

Answer 1

\[\int\limits_{}^{}\sqrt{1+\sin2x+(\frac{ \cos2x }{ \sqrt{1+\sin2x} })^{2}}dx\]
Hard to tell if that was what you meant or not.

Answer 2

o my bad its 0<x<pi*rad2

Answer 3

and yes that is what i mean

Answer 4

Alright, I got it, lol

Answer 5

So that 2nd power distributes and gives us:
\[\int\limits_{}^{}\sqrt{1+\sin2x+\frac{ \cos ^{2}2x }{ 1+\sin2x }}dx\]
Now that cos^2 Im going to change using the pythagorean identity:
\[\int\limits_{}^{}\sqrt{1+\sin2x+\frac{ 1-\sin ^{2}2x }{ 1+\sin2x }}dx\]
Now I can take that 1 - sin^2 and factor it like a difference of squares:
\[\int\limits_{}^{}\sqrt{1+\sin2x+\frac{ (1-\sin2x)(1+\sin2x) }{ 1+\sin2x }}dx\]

Im sure you can get it from there : )

Answer 6

thx again! :D

Answer 7

Lol, sure xD