Question

Hi. What is the simplest solution in integrating sqrt(sin^2 (2x) + 4cos^2 (2x)) dx ?? Thanks

Answer 1

Don't think there is a simple solution to be honest. I have something, but not 100% sure it's....legal to do the substitution I did.

Answer 2

Are you really sure this is what you mean? What level math course are you in? (It's not an easy problem.)

Answer 3

If there was a "4" in front of the sin^2(2x), then it would be easy!

Answer 4

Well yeah, lol. Kinda funky, but I did a trig substitution.....even though there's already a trig function in there. That's the part I'm unsure if it's legal to do, lol. It gets an answer xD

Answer 5

The answer to the problem, as stated, involves an elliptic integral! (Go to http://www.wolframalpha.com, and type in something like this:
integral of sqrt(sin^2(2x) + 4cos^2(2x)) dx

Answer 6

Never seen such a thing, lol. Guess that means it's ompossible to do my method x_x

Answer 7

I guess it's really unintegrable. But still we can get a value if there's limit given through series of integration.

Answer 8

Depends on whether or not it's invalid to do this:
\[\int\limits_{}^{}\sqrt{\sin ^{2}(2x)+4\cos ^{2}(2x)}=\int\limits_{}^{}\sqrt{1+3\cos ^{2}(2x)}dx\]
Let:\[\sqrt{3}\cos(2x)=u=\tan \theta \]and let:
\[\sqrt{1+3\cos ^{2}(2x)}=\sec \theta\]
\[\int\limits_{}^{}\sec \theta du \]
If u = tan(theta) then du = sec^2(theta) leaving:
\[\int\limits_{}^{}\sec ^{3}\theta d \theta\]
This was the direction I was heading in. Integration by parts, do back-substitution in the end.

Answer 9

hey people I can't understand the solution i found on wolframalpha.com. Why are there E and Round? What do those things mean?? O.O

Answer 10

Something above our level I guess, lol. Elliptic integral. Do you kinda see what I was trying to do with your integral?

Answer 11

Well maybe you're right about that something above our level but I want also to know that. Nyahaha. Yeah, because we did the same.

Answer 12

Lol, I see. Yeah, Id probably just go with that. You get:
\[\frac{ 1 }{ 2 }\sec \theta \tan \theta+\frac{ 1 }{ 2 }\ln|\sec \theta+\tan \theta|\]
\[\frac{ 1 }{ 2 }(\sqrt{1+3\cos ^{2}(2x)}(\sqrt{3}\cos2x)+\frac{ 1 }{ 2 }\ln|\sqrt{1+3\cos ^{2}(2x)}+\sqrt{3}\cos(2x)|+C\]