Trigonometric substitution and triangles

Question

I'm learning trigonometric substitutions and am having a bit of trouble understanding the intuition behind the conversions (why do most use secant?). If you could explain the conversions geometrically using a triangle, that would be very helpful. For example, if we have $$\int \frac{\sqrt{x^2-4}}{x}\,dx$$ I tried to construct a triangle like so:

To get $$\sin(\theta)=\frac{\sqrt{x^2-4}}{x}$$

But this is incorrect for use as a substitution. Why?

Answer 1

The intuition to use trig-sub is because sometimes a problem becomes easier when you make substitutions. Mainly, grouping "nasty" looking quantities into neat little packages can save you a headache. Notice that where you placed the quantities "$2$" and "$\sqrt{x^2-4}$" in your picture were arbitrary. Try switching them and see what happens. Your answer will still be correct.

Answer 2

I think the way to think about trigonometric substitutions is to remember the fundamental result that

$$\sin^2a + \cos^2a = 1$$

for any angle $a$. So, in your example, you have a square-root

$$\sqrt{x^2 - 4}$$

Imagine for a moment that that 4 was a 1 and that they were reversed, like so:

$$\sqrt{1 - x^2}$$

Then the substitution $x = \sin\theta$ would result in something simple, without the square-root, namely

$$\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \cos\theta$$

That's the gist behind trig substitutions.

Now, in your case in particular, things aren't so simple so let's work it out step by step. First, imagine the 4 is a 1:

$$\sqrt{x^2 - 1}$$

We can't use $x = \sin\theta$, nor $x = \cos\theta$, here because we'd get the square-root of a negative quantity. For instance, using $x = \sin\theta$,

$$\sqrt{x^2 - 1} = \sqrt{-\cos^2\theta}$$

So how can we use the fundamental relation at the top of this post to get something like $x^2 - 1$? Well, divide both sides by $\cos^2a$ and write it as follows:

$$\frac{\sin^2a}{\cos^2a} + \frac{\cos^2a}{\cos^2a} = \frac{1}{\cos^2a} \quad \rightarrow \quad \tan^2a = \frac{1}{\cos^2a} - 1 = \sec^2a - 1$$

Now it's obvious that $x = \sec\theta$ give us

$$\sqrt{x^2 - 1} = \sqrt{\sec^2\theta - 1} = \sqrt{\tan^2\theta} = \tan\theta$$

and we once again got rid of the square-root. The only thing that remains is the 4. We can deal with that by scaling. Choose $x = 2\sec\theta$, instead, so

$$\sqrt{x^2 - 4} = \sqrt{4\sec^2\theta - 4} = \sqrt{4(\sec^2\theta - 1)} = \sqrt{4\tan^2\theta} = 2\tan\theta$$

So, to summarise, all you really need to remember is the fundamental relation at the top of this post and "massage" it as necessary to get rid of any square-roots that are in your way.

The two main forms that you'll use are:

$$\sin^2a + \cos^2a = 1$$

and

$$\tan^2a + 1 = \sec^2a$$

which, as we've seen, are really one and the same.

Answer 3

The idea is to eliminate the square root. For $\sqrt{x^2+a^2}$, the substitution $x=a\tan t$ makes the square root

$$\sqrt{a^2\tan^2t+a^2}=\sqrt{a^2\sec^2t}=\pm a\sec t$$

Similarly, $x=a\sin t$ eliminates the square root for the form $\sqrt{a^2-x^2}$. Your example is the third case, $\sqrt{x^2-a^2}$. So this time secant eliminates the radical.

$$x=2\sec t,dx=2\sec t\tan tdt$$

The integral now becomes

$$\int\frac{\sqrt{4\sec^2t-4}}{2\sec t}2\sec t\tan tdt=$$ $$\int\pm2\tan^2tdt$$

One more trig identity and the integral is easily solved. I believe the sign issue is usually ignored, though it may come into play for definite integrals based on the bounds.

Answer 4

It's not that it's wrong to take on the integral the way you propose -- it's just not awfully convenient. If we implicitly differentiate your substitution equation, we obtain

$$ \cos \theta \ d\theta \ = \ \frac{x \ \cdot \frac{1}{2} \ (x^2 - 4)^{-1/2} \ \cdot 2x \ - \ (x^2 - 4)^{1/2} \ \cdot \ 1}{x^2} \ dx $$ $$ = \ \frac{4}{x^2 \ (x^2 - 4)^{1/2}} \ dx \ = \ \left(\frac{2}{x} \right)^2 \cdot \ \frac{2}{ \ (x^2 - 4)^{1/2}} \ \cdot \ \frac{1}{2} \ \ dx $$ $$ = \frac{1}{2} \ \cdot \ \cos^2 \theta \ \cdot \ \frac{1}{\tan \theta} \ \ dx $$

[using your right triangle]

$$ \Rightarrow \ \ dx \ = \ 2 \ \tan \theta \ \cdot \ \frac{\cos \theta}{\cos^2 \theta} \ \ d\theta \ = \ 2 \ \tan \theta \ \sec \theta \ \ d\theta \ \ , $$

which is just the result Mike shows. Following your lead, the integral becomes

$$ \int \ \sin \theta \ ( \ 2 \ \tan \theta \ \sec \theta \ \ d\theta \ ) \ = \ 2 \int \ \sin \theta \ \tan \theta \ \frac{1}{\cos \theta} \ \ d\theta \ = \ 2 \int \ \tan^2 \theta \ \ d\theta \ \ . $$

There are easier ways to get there, though.