Trig. Substitution – Integration

Q:  ∫1 / (u²+ 2) du

A:  When looking at this problem, I think the first step is trig. substitution.  Why do I think that?  Experience.

I remember the two most important trig identities when considering trig substitution in intergration:

(1) sin²x + cos²x = 1

(2) tan²x + 1 = sec²x

Since I see u² + 2 in my original problem, I think that this looks more like (2)… Therefore, I try the following:

Let u = √(2)*tanx

Why did I also include √(2)?? Watch and see:

If u = √(2)*tanx then du = √(2)*sec²x dx

Now, substitute in u and du into our orignal equation, like so:

∫1 / (u²+ 2) du

∫1 / ((√(2)*tanx)²+ 2) [√(2)*sec²x dx]

Do the colors help to show where I have made the substitutions, or do they just make it more confusing?

Anyhow, now for some house keeping (square things out, multiply my fractions, etc…):

∫1 / ((√(2)*tanx)²+ 2) [√(2)*sec²x dx] = ∫1 / (2*tan²x+ 2) [√(2)*sec²x dx]

= ∫√(2)*sec²x / (2*tan²x+ 2) dx

Now, I have one fraction.  In the bottom, notice both constants are “2s”?  Thanks to my intial choice of √(2)!  So, factor out a 2:

∫√(2)*sec²x / [2(tan²x+ 1)] dx

Wait… Remember that trig identity from above? tan²x + 1 = sec²x

Therefore, replace tan²x + 1:

∫√(2)*sec²x / (2sec²x) dx

Aha.  Now the secants cancel!

∫√(2)/2 dx

We are just left to integrate a constant:

∫√(2)/2 dx = √(2)/2 * (x) + C

Well, we are almost done.  Since the orignal problems were given to us in terms of “u”, we better put it back into u’s and not x’s.

Recall: u = √(2)*tanx.  Solve for x:

u = √(2)*tanx

u / √(2) = tanx

arctan [u / √(2)] = x

Therefore, our answer: √(2)/2 * (x) + C = √(2)/2 * (arctan [u / √(2)]) + C

10 thoughts on “Trig. Substitution – Integration

  1. We have not even covered trig sub yet, I’ll post the work I did and lets see if we can work through it, maybe the answer I have is the same I don’t know, I’ll do that in a minute.


  2. This is the work that I did I got close to the right answer but of course close is only good in horseshoes and hand grenades lol.

    Integrate {(1)/(x^2+2x+3)} dx

    Check discriminant: b – 4ac <= 0 than complete square
    Discriminant = -10 so complete the square

    (x+1)^2 + 2 is the new denominator

    Let u = x + 1
    so du = dx

    We then get the following:
    Integrate {1/[(x + 1)^2 +2]} dx
    = Integrate {1/(u^2 + 2)} du

    Int. {1/2(u^(2)/2 + 1)}
    = 1/2 Int. {1/{(u/ radical 2)^2 +1]}

    = sqrt(2)/2 Int. {1/(u^2 + 1)}
    = sqrt(2)/2 arctan(u) + constant
    = sqrt(2)/2 arctan(x+1) + constant

    Above is the work I did for the same problem that you did above, are they same and is what I did remotely correct?


  3. Everything in BOLD up until the “—————————” I added is good (I edited your comment)

    The very next line, where a sqrt(2) is factored out has gone wrong.


    • I took like this:
      (U/sqrt(2))^2 and moved the radical 2 from under the U to the numerator of the 1/2 outside the radical sign is that incorrect? I tried to follow some notes that I had from class


  4. I will pick up where things went south:

    = 1/2 Int. {1/{(u/ radical 2)^2 +1] du }

    Immediately from here, we can see that “arctan” will be in our answer….

    Now, another substitution. Let w = u/sqrt(2), dw = 1/sqrt(2) du
    So, du = sqrt(2) dw

    = 1/2 int { 1 / (w^2 + 1) sqrt(2) dw}

    Clean house:
    = sqrt(2)/2 int { 1 / (w^2 + 1) dw}
    = sqrt(2)/2 * arctan(w) + C

    = sqrt(2)/2 * arctan(u/sqrt(2)) + C


    • Aaahhhhhh I see another substitution was needed to do it correctly all I was missing was that stinkin radical 2 in the denomicator of the arctan always the small stuff lol. thanks so much Stacy, this is gettin tweeted.


  5. Response to comment #4:
    Yes, this is incorrect because you are factoring out something from a denominator and placing it in a numerator — it just changes spots too magically. Also, since the [(u/sqrt2)^2 + 1] is bound together by a plus sign, you must be factoring numbers out of BOTH of them in order to pull anything out…. Make sense?


  6. yeah totally makes sense I will have to go through my notes again because I don’t remember doing anything like that, maybe the prof. made a mistake? lol that would be extra points on the exam haha


  7. My first thought was just to use the arctan integration formula, with a=sqrt(2). It works a little (a lot) quicker – but now that I’m also teaching calculus 2 it’s nice to see how trig substitution will also work!


  8. Hi Michelle,
    Yes, using the arctan integration formula is great if you either have it memorized or are able to use a book for reference. But, in the event that this shows up on a test and you have to figure it out from scratch it is very nice to know the trig substitution method or the other substitution methods!


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