how can i solve integral 1/square root of (sin4x)?

Answer 1

Sorry, Hellogood...'s answer is WRONG!
One CANNOT use the power rule for transcendental functions!!
Doug has the right idea. It is not in the form of an elliptic
integral but it can be changed into one by substitutions.
Let's first get rid of the 4:
Let u = 4x, x = u/4, dx = 1/4*du
So we have 1/4*∫( du/√sin u)
Now let t = sin u, u = arcsin t, du = 1/√(1-t^2).
So we finally get
1/4* ∫ dt/√t(1-t²).
Since we have the square root of a cubic,
we now have an elliptic integral.
BTW, I ran your original integral through integrals.wolfram.com
and got
-1/2*F(1/4π - 2x| 2),
where F is an elliptic integral of the first kind.

Answer 2

Okay, you have 1srt(sin4x). Firstly put this in index form which is (sin4x)^1/2.
Add 1 to the power so u get (sin4x)^3/2.
Then you know you must do (3/2) divided by something to equal 1 (the coefficent of your original function. This is (2/3).
So your answer is (2/3(sin4x)^3/2) + k. .u have to +k as this is a indefinate integral (no x values are defined). k is just a constant. this can be checked by differentiating ur answer, u should get (sin4x)^1/2. ive done this to check what i have told u is correct and it is.

Answer 3

I think you better go back and look at that integral again. It looks very much as if it's a 'standard form' elliptic integral of the first kind. Or are you in a graduate class in ODE's or Analysis and working on elliptic integrals?


Doug

Answer 4

try a trig substitution or the CRC integral tables