What is the integral of f'(x)=sin^-2(7x^3/2x^-2) with respect to x?

Question

What is the integral of f'(x)=sin((7*x^3)/(2*x^(-2)))^(-2) with respect to x? You need not post how to do the solution, unless you want to be super helpful. I've gotten a lot of different answers, and someone told me that there's some gamma thing going on and branch cuts in the complex z plane and this wierd stuff.

I don't know why it did the ... in the question, I clearly put -2 but it should be csc^2.

That last extension might not have been very clear. I'm looking for ∫csc[(7*x^3)/(2*x^[-2])]^2 dx.

I don't think this will be easy to solve by hand, I'd use mathmatica, but the computer with that on it is several hundred miles away.

This also seems to be too difficult to solve with web-based integration websites.

Answer 1

It's definitely nonelementary.
First, simplify the inside to get
∫ csc^2(7/2*x^5)dx
Now let u = 7/2 x^5
dx = Cu^(-4/5) du,
where C = (2/7)^(1/5)*1/5.
Finally integrate by parts. I'll leave out the gory details,
but it boils down to integrating
∫u^(-9/5)cot u du,
and here we reach a dead end.
There seems to be no way to calculate this
last integral. Sorry!

Answer 2

cos(2x) = 1 - 2 sin^2(x) Therefore cos(2 * 3/4 x) = 1 - 2 sin^2(3/4 x) cos(3/2 x) = 1 - 2 sin^2(3/4 x) 2 sin^2(3/4 x) = 1 - cos(3/2 x) sin^2(3/4 x) = 1/2 - (1/2) * cos(3/2 x)] The integral = x/2 - (1/2) * (2/3) * sin(3/2 x) + C = x/2 - (1/3) * sin(3/2 x) + C (where C is constant of integration)

Answer 3

BRO. the answer is a bit critical what i can help you out is by telling you a way to solve the ques. what you need to do is to use a formula i.e. ( U.V ) or
U x ∫ V d dx - ∫ ( U d x V ) dx
dx dx

where U is your actual question and V is 1

Answer 4

You're making to work my brain, jejeje. I'm trying to do it via complex integration but I can't find the solution yet. I'll keep trying.

Answer 5

hello
any integration problem you want to solve heres a free link that solves everysingle impossible integrals

http://integrals.wolfram.com/index.jsp