Finding the Anti-Derivative of x*cosh(x^2) using Hyperbolic Identities

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The discussion focuses on finding the anti-derivative of x*cosh(x^2) using integration techniques. The initial approach involved integration by parts, but the challenge arose with the integral of cosh(x^2). Participants suggested using u-substitution, where u = x^2 and du = 2x dx, simplifying the integral to (1/2)∫cosh(u)du. The cancellation of variables was highlighted as a key step in solving the problem. Ultimately, the solution was reached through u-substitution, demonstrating its effectiveness over integration by parts.
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Homework Statement



Find the anti derivative of \int xcosh (x^2) dx

Homework Equations



By parts formula and Hyperbolic Identities of sinh x and cosh x as well as others

The Attempt at a Solution



\int xcosh (x^2) dx

The problem I'm having is integrating \int cosh (x^2) dx

I tried setting variables u=x and \frac{dv}{dx}= \int cosh (x^2) dx with the assumption this could be solved using the by parts formula.

I then concentrated specifically on solving \int cosh (x^2) dx. I haven't found a method that I know of that's appropriate given that the composite is (x^2) and not (cosh x)^2. Wolfram Alpha shows the solution with an error function - which I know nothing about yet.

I've touched up on Euler's formula cosx+isinx=e^{ix} and its parallel sinhx+coshx=e^x and I'm just about to learn its applications, maybe it should be used here. This area is new to me so light explanations are wise at this time.
 
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You don't need parts, all you need is u-substitution. u = x^2 and du = 2x dx
 
Panphobia said:
You don't need parts, all you need is u-substitution. u = x^2 and du = 2x dx

I tried that but gave up because of the extraneous x which would mean substituting it for \sqrt {u}.

The x in the numerator cancels ut the out in x in the denominator.

Sorry I got it.
 
look
u = x^2
du = 2x * dx

du/2 = x * dx

(1/2)∫cosh(u)du

Now from there its pretty easy as you can see.
 
Panphobia said:
look
u = x^2
du = 2x * dx

du/2 = x * dx

(1/2)∫cosh(u)du

Now from there its pretty easy as you can see.


Yes the x variables cancel each other out. I figured it out
 
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