Integral with trig substitution

[howsockgothap]

Homework Statement

Using trig substitution integrate x²/(x⁴+6x²+9)

Homework Equations

The Attempt at a Solution

I'm almost completely positive I need to use partial fraction decomposition... doing this I end up having the integral of 1/(x²+3)-3/(x²+3)²

There aren't any square roots in that, so I don't understand how to set up my trig substitution.

 

[issacnewton]

Use the substitution

x = \sqrt{3} \tan \theta

 

[Dick]

You want to write the sum x^2+3 as a single term. How about x=sqrt(3)*tan(t)?

 

[howsockgothap]

Ok working that out I get sqrt3sec²(t)/(sqrt3sec(t)) - 3sqrt3sec²(t)/3sec²(t)...


so by this it seems I need to take the integral of sec(t)-sqrt3 dt

Does this seem correct?

 

[Dick]

Ok working that out I get sqrt3sec²(t)/(sqrt3sec(t)) - 3sqrt3sec²(t)/3sec²(t)...


so by this it seems I need to take the integral of sec(t)-sqrt3 dt

Does this seem correct?

It would seem correct if I intentionally try to make a ton of algebraic mistakes. Can you show what you did to get there so that people can help you correct them?

 

[issacnewton]

Yes, follow what Dick is saying and learn some Latex so that equations are readable . Its very easy to master

 

[Dick]

My main message is that the content is wrong. The main problem is that you didn't show how you got there, so what you are doing wrong is anyones guess. isaacNewton is right, that latex would make it easier to read. But it's not essential. Just show how you got there.

 

[jambaugh]

Some general advice of trig substitutions:

The use of trig subst. applies to any of the three forms:
x^2 + a^2, \quad x^2 - a^2, \quad a^2 - x^2
where you need to convert this to a single factor.

The applicable identities are the various forms of the Pythagorean identity:
\cos^2(t) + \sin^2(t) = 1\leftrightarrow 1 - \sin^2(t)=\cos^2(t)

\frac{\cos^2(t) + \sin^2(t) = 1}{\cos^2(t)}\to 1 + \tan^2(t) = \sec^2(t) \leftrightarrow \sec^2(t) - 1 = \tan^2(t)

and similarly dividing by sine squared but this gives identities with no further utility than what we have already.

One applies the appropriate substitution:
x^2 + a^2 \to x = a \tan(t)
x^2 - a^2, \to x = a \sec(t)
a^2 - x^2 \to x = a \sin(t)

and this gives a single term instead of a sum of two squared terms... which may or may not be useful in integration but typically gives a useful path to the appropriate anti-derivative. It is not always the case that one is resolving a radical and one can usually carry out an equivalent non-trigonometric algebraic manipulation to accomplish the same task. However such alternatives typically require a great deal of insight, luck, and/or experience while the trig substitutions allow for systematic attack.

You may, for example carry out the appropriate trig substitution prior to partial fraction decomposition, and indeed it may even save you some work later. Partial fraction decomposition of trig-substitutable expressions have parallel trig identities which may be applied but needn't be if the original integrand is simpler.

Finally don't forget those differential factors in the substitution, you wish to integrate: \frac{x^2}{x^4+6x^2+9} dx
not just
\frac{x^2}{x^4+6x^2+9}

 

[howsockgothap]

If x = sqrt3 tan(t) then dx=sqrt3 sec²(t) and x²+3=sqrt3 sec(t)

I plugged that into 1/(x²+3)dx -3/(x²+3)²dx

so: (sqrt3 sec²(t))/(sqrt3 sec(t)) dt - 3sqrt3sec²(t)/3sec²(t)

I canceled out everything I could to get sec(t)-sqrt3 dt...


which is wrong, obviously. I guess that should instead be that sqrt(x²+3)=sqrt3 sec(t)

 

[issacnewton]

one of the mistakes found...in the first line, it should be

dx= \sqrt{3} \sec^2(t) \,\, dt

 

[Dick]

If x = sqrt3 tan(t) then dx=sqrt3 sec²(t) and x²+3=sec(t)

I plugged that into 1/(x²+3)dx -3/(x²+3)²dx

so: (sqrt3 sec²(t))/(sqrt3 sec(t)) dt - 3sqrt3sec²(t)/3sec²(t)

I canceled out everything I could to get sec(t)-sqrt3 dt

x^2+3 isn't sec(t). It's 3*sec(t)^2, isn't it? Showing what you did does help.

 

[howsockgothap]

Right, so fixing both of those mistakes I work it down to 1/sqrt3 - sqrt3/3sec²(t)

since it becomes (1/(3sec²t) - 3/(9sec⁴t)) sqrt3sec²tdt


and at least according to my messed up math that just leaves me with t/sqrt3 - the integral of (3/sqrt3) cos²t

 

[Dick]

Right, so fixing both of those mistakes I work it down to 1/sqrt3 - sqrt3/3sec²(t)

since it becomes (1/(3sec²t) - 3/(9sec⁴t)) sqrt3sec²tdtand at least according to my messed up math that just leaves me with t/sqrt3 - the integral of (3/sqrt3) cos²t

Your messed up math is getting less messed up by the minute. Now you just have to integrate cos(t)^2. And then find that expression in terms of x. If you really need to go back to x.

 

[howsockgothap]

Alright, so I use the double angle formula and the integrand becomes -sqrt3/6 (1+cos2t)

Plus of course t/sqrt3 from earlier

Then I just work that out, and I ended up with (sqrt3/6)(t+sin2t)

And if that's right, I think I've got it from there.

 

[howsockgothap]

Found a mistake in my own work already. Should be -sin2t

 

[Dick]

Alright, so I use the double angle formula and the integrand becomes -sqrt3/6 (1+cos2t)

Plus of course t/sqrt3 from earlier

Then I just work that out, and I ended up with (sqrt3/6)(t+sin2t)

And if that's right, I think I've got it from there.

I think you might be off by maybe a factor of 2 and a sign. Might want to check it again. But you've got all the right ideas now, so I'm as likely to have made a mistake as you are.

 

[Dick]

Found a mistake in my own work already. Should be -sin2t

Bravo. It's just mechanics now.