Integration question relating to inverse trig functions.

**Ant** · January 4th 2011, 05:50 PM

Hi,

I've been doing some differentiation of inverse trig. In doing this I had to find the derivative of:

$<br /> arctan (3x/2)$

I found this to be: $6/(9x^2 + 4)$

To check this answer I thought I would try to integrate to see if I arrived back at the original question. However, I'm a bit stumped as to how I should go about integrating it?

I know that the original derivative was derived using the chain rule so is there some substitution that I can use to find the integral of $6/(9x^2 + 4)$

Thanks,

**snowtea** · January 4th 2011, 05:53 PM

Your question asked for the derivative of
y = arctan(3x/2)

so x = 2tan(y)/3

Use this substitution for the integral. It should simplify very nicely (assuming you did your differentiation correctly).

**pickslides** · January 4th 2011, 05:59 PM

Commit this to memory $\displaystyle \int \frac{a}{a^2+x^2}~dx = \text{Tan}^{-1}\frac{x}{a}+C$

**Ant** · January 4th 2011, 06:01 PM

Thanks for the quick reply.

How would you find the right substitution if you were just given the question:

Integrate $6/9x^2 +4$

I'm sure that I could spot that it related to arctan (the x^2 and the positive constant is the give away) but how could I get it into a form from which I could integrate?

Sorry if I'm missing the obvious...

**Ant** · January 4th 2011, 06:07 PM

Got it now.

If I compare $6/9x^2 +4$

With

$\displaystyle \int \frac{a}{a^2+x^2}~dx = \text{Tan}^{-1}\frac{x}{a}+C$

Divide through by 9 to make the coeff of x^2 equal to 1. Immediately you have the general form given above.

Thanks

**snowtea** · January 4th 2011, 06:10 PM

It is actually standard to try to substitute trig functions when you see forms like $ax^2 + b$ in the denominator of an integral (without an $x$ term in the numerator).
Normally you use the trig function that allows you to simplify the expression with a trig identity.
For example when you see $1 - x^2$ in the denominator, you might use $x = sin(y)$ .
There may be multiple trig subs (or even hyperbolic subs, i.e. sinh, cosh) that work, but they should all come out to be equivalent answers.

**Ant** · January 4th 2011, 06:22 PM

I can see how you could reasonably easily work out which trig or hyperbolic to sub in. But given the integration question above, all I would have been able to tell you was that I should sub in tan. I wouldn't have been able to get the rest of the substitution at all...

**snowtea** · January 4th 2011, 06:29 PM

Sure you can the 9 and 4 have something to do with the 3 and 2

Normally, if you have $(a + bx^2)$ you think $a(1 + \frac{b}{a}x^2)$ .
Now the coefficient should cancel out $\frac{b}{a}$ .

**pickslides** · January 4th 2011, 06:35 PM

You can also find the result on your own if you don't remember the derivative/identity

$\displaystyle \int \frac{6}{9x^2+4}~dx$

Make $\displaystyle x = \frac{2}{3}\tan\theta \implies \theta = \tan^{-1}\frac{3x}{2}$ and $\displaystyle dx=\frac{2}{3}\sec^{2}\theta d\theta$

$\displaystyle \int \frac{6}{9( \frac{2}{3}\tan\theta )^2+4}\times \frac{2}{3}\sec^{2}\theta d\theta$

$\displaystyle \int \frac{6}{9\times \frac{4}{9}\tan^2\theta +4}\times\frac{2}{3}\sec^{2}\theta d\theta$

$\displaystyle \int \frac{6}{4\tan^2\theta +4}\times\frac{2}{3}\sec^{2}\theta d\theta$

$\displaystyle \int \frac{6}{4(\tan^2\theta +1)}\times\frac{2}{3}\sec^{2}\theta d\theta$

$\displaystyle \int \frac{6}{4\sec^2\theta }\times\frac{2}{3}\sec^{2}\theta d\theta$

Cancelling down

$\displaystyle \int 1 ~d\theta = \theta +C = \tan^{-1}\frac{3x}{2}+C$

**Ant** · January 4th 2011, 07:12 PM

Thanks for all the replies. They've been extremely helpful.

My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?

I've tried just subing in x = tan y. In the above example as a test to see if it's really always necessary to include in the correct coefficients and I haven't managed to make it work yet. (which may well be because my algebra is somewhat lacking...)

**TheCoffeeMachine** · January 4th 2011, 07:19 PM

Originally Posted by Ant

My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?

No, but with enough practice, you'll be able to spot the relevant substitution (at least for the regular ones).

**pickslides** · January 4th 2011, 07:22 PM

Originally Posted by Ant

My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?

Practice makes perfect here, the more you do the better you will get at identifying which substitution to use. Trial and error won't hurt, patterns will present.

Look here for more info

Trigonometric substitutions

Originally Posted by Ant

I've tried just subing in x = tan y. In the above example as a test to see if it's really always necessary to include in the correct coefficients and I haven't managed to make it work yet. (which may well be because my algebra is somewhat lacking...)

Maybe this is the case, once again practice will make perfect, you are looking to cancel any coeffecients leaving something neat like $1-\cos^2\theta, 1-\sin^2\theta , 1+\tan\theta$

**tom@ballooncalculus** · January 5th 2011, 01:37 AM

Originally Posted by Ant

My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?

I've tried just subing in x = tan y. In the above example as a test to see if it's really always necessary to include in the correct coefficients and I haven't managed to make it work yet.

Instead of assuming you need a sub in the form

x = some function of tan theta,

you might prefer to get the denominator in the form 1 + some square or other, thereby making it analogous to the left hand side of $1 + \tan^2\theta = \sec^2\theta$ , which is the whole point of the sub. Then it's obvious to sub tan theta for directly for the square you made.

Just in case a picture helps...

Then adjust for this expression being a derivative via the chain rule

... where (key in spoiler) ...

Spoiler:

Then you can make the direct analogy with integrating 1 + tan squared...

See also http://www.mathhelpforum.com/math-he...tml#post546765

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