integration problem!!?

Question

integral of 1/(-x * inverse sin x) dx

Answer 1

I don't think this function has an integral that can be written out using familiar functions.

I'll use "arcsin x" to mean the inverse sine of x. I'll use "x^(1/2)" to mean the square root of x.

Use the substitution u = arcsin (x). Then du = 1/(1 - x^2)^(1/2) dx.

Since u = arcsin (x), then x = sin (u).

We have du = 1/(1 - x^2)^(1/2) dx. Plugging in x = sin (u) gives

du = 1/(1 - sin^2 u)^(1/2) dx.

Now apply the identity 1 - sin^2 u = cos u. We get

du = 1/(cos^2 u)^(1/2) dx.

Now, (cos^2 u)^(1/2) = cos u, so we have

du = 1/cos u dx

so

cos u du = dx.

Now we know u = arcsin(x), x = sin (u), and cos u du = dx, so we can substitute into the original integral, giving us

∫(1/(-sin (u) * u) * cos u du)

Which equals

∫(-cot u)/u du.

To the best of my knowledge, this is an "impossible" integral to compute. This doesn't mean an antiderivative doesn't exist; it just means that the antiderivative isn't one that we can easily express using well-known functions. (For example, neither my calculator nor the integrator at http://integrals.wolfram.com/index.jsp can do this indefinite integral.)

Answer 2

Rewrite this as a ODE, y'= 1/(-x arcsin x), now expand the Taylor series and differentiate the terms and you can reduce it to an infinate series which converges to the solution for all x, thats the brute force method if some elementary function is not availible. Im not going to do the work for you tho cause i dont even understand the point of askign this question lol.

Answer 3

Is there even an answer for that? I don't see any possible way of integrating it. All I got so far is moving the negative sign out of the integrand, lol. From there, I can only think of integration by parts but does that work when you have 1/(xarcsinx)?

Answer 4

Integral ( 1/ (-x arcsin(x) ) dx

I don't believe this has a solution with the elementary functions we know. But maybe someone can prove me wrong.

Did you write it correctly?

Answer 5

So are you a huge fan of math or what? transform your question a little and I will try and give you an equation.

Answer 6

crucial by using areas is: ?udv=uv-?vdu so after substitution and placing u=x and dv=F'(x), ?xF'(x)dx=xF(x)-?F(x) and because ?F(x)=G(x) because of fact F(x)=G'(x) ?xF'(x)dx=xF(x)-G(x).