What substitution to use in this type of integral?

In summary: I was wondering how he arrived at his last formula...In summary, the conversation discusses different attempts at solving the integral of a complex function and suggests using the tangent half-angle substitution or completing the square and using a linear translation to simplify the integral. The conversation concludes with a discussion on how to arrive at the final formula using trigonometric identities.
  • #1
gruba
206
1

Homework Statement


Find the integral [itex]\int \frac{3x+1}{(x^2-x-6)\sqrt{3x^2+4x-7}}\mathrm dx[/itex]

2. The attempt at a solution
I have tried the types of substitutions of irrational functions, and Euler substitutions.
However, it seems that nothing simplifies this integral.

What substitution is useful for this type of integrals?
 
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  • #2
gruba said:

Homework Statement


Find the integral [itex]\int \frac{3x+1}{(x^2-x-6)\sqrt{3x^2+4x-7}}\mathrm dx[/itex]

2. The attempt at a solution
I have tried the types of substitutions of irrational functions, and Euler substitutions.
However, it seems that nothing simplifies this integral.

What substitution is useful for this type of integrals?
On this one, I would try completing the square of the polynomial inside the square root sign in the denominator. Once you do that, possibly a linear translation to get this part as sqrt(z^2-1). Then let ## z=sec(\theta) ##. I don't know that this will work, but it might be worth a try. This one looks complex in any case.
 
  • #3
gruba said:

Homework Statement


Find the integral [itex]\int \frac{3x+1}{(x^2-x-6)\sqrt{3x^2+4x-7}}\mathrm dx[/itex]

2. The attempt at a solution
I have tried the types of substitutions of irrational functions, and Euler substitutions.
However, it seems that nothing simplifies this integral.

What substitution is useful for this type of integrals?
Try starting with ##u=3x+1##. That turns out to make the inside of the radical relatively nice.
 
  • #4
vela said:
Try starting with ##u=3x+1##. That turns out to make the inside of the radical relatively nice.
Can you double-check this one and/or explain further? This problem stumped me as well. I tried my solution and it doesn't seem to work either because mine results in an integration of a couple of terms of the form ## du/(A+B\sec(u)) ## which I don't have an answer for.
 
  • #5
Charles Link said:
Can you double-check this one and/or explain further? This problem stumped me as well. I tried my solution and it doesn't seem to work either because mine results in an integration of a couple of terms of the form ## du/(A+B\sec(u)) ## which I don't have an answer for.
https://en.wikipedia.org/wiki/Tangent_half-angle_substitution
 
  • #6
Charles Link said:
Can you double-check this one and/or explain further? This problem stumped me as well. I tried my solution and it doesn't seem to work either because mine results in an integration of a couple of terms of the form ## du/(A+B\sec(u)) ## which I don't have an answer for.
I didn't actually work it out. I just tried the "obvious" substitutions based on the terms in the integrand and noticed that one made the stuff inside the radical look nice.

Anyway, I futzed with it a bit more, and it looks like you can get it down to an integrand (I left out some overall constant factors) which looks like
$$\frac{5-\cos\theta}{(5-11\cos\theta)(5+4\cos\theta)}\,d\theta,$$ at which point you can proceed with the technique micromass pointed to.
 
  • #7
I worked through it a little further using the ## z=sec(\theta) ## mentioned in my first response. With some partial fractions, I got it to the form of ## du/(A+B\sec(\theta) ##. (Two terms of this type.) (I don't get the cosines that vela has.) The ## z=tan(x/2) ## substitution micromass suggested works on these terms to get something that looks like ## du/(1+u^2) ## which integrates to inverse tangent. (I had previously seen that substitution many years ago when we used the Thomas Calculus text.) Thank you both for your inputs. The OP doesn't seem to be responding, but it is an interesting calculation. I'm currently retired, and simply doing it as a hobby... A follow-on: I see to get vela's cosine terms from terms with secants, all that is necessary is to multiply numerator and denominator by ## cos(\theta) ##...
 
Last edited:
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1. What is substitution in integration?

Substitution, also known as the change of variable method, is a technique used in integration to simplify a complex integral by replacing the variable with a new one. This new variable is chosen in such a way that it transforms the integral into a more manageable form.

2. When should I use substitution in integration?

Substitution is typically used when the integral has a complicated integrand, or when it involves a composition of functions. It is also useful when the integral involves a trigonometric function or a radical expression.

3. What is the general process of substitution in integration?

The general process involves choosing a suitable substitution variable, substituting it into the integral, simplifying the integral in terms of the new variable, and then integrating with respect to the new variable. Finally, the result is expressed in terms of the original variable.

4. How do I choose the appropriate substitution variable?

Choosing the appropriate substitution variable can be done by identifying a part of the integrand that closely resembles a derivative of the variable. This variable should also be easy to integrate and should simplify the overall integral.

5. Are there any common substitution formulas that I should be aware of?

Yes, there are several common substitution formulas, such as u-substitution, trigonometric substitution, and hyperbolic substitution. It is important to familiarize yourself with these formulas and their applications to effectively use substitution in integration.

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