# Trigonometric substitution - Why?

• Runei
In summary: So, cos theta and arccos x are both values of x within the triangle and can be substituted for one another without confusion.However, when I looked at the example on sponsoredwalk's website \int_{a}^{b}\frac{1}{\sqrt{1-x^2}}dx I found that the substitution x= a sin(\theta) does not appear to work \int_{a}^{b} \ csc( \theta ) \ dx GraphicsIn summary, Trigonometric substitution - Why?
Runei
Trigonometric substitution - Why?

Hey guys

Im sitting here with trigonometric substitution problems, and I have a kind of a problem.

I can't see WHY it is legal to substitute x for a sin ($$\theta$$)

If you have a the integral:

$$\int$$$$\frac{1}{\sqrt{1-x^2}}$$dx

Then I know the substitution would be

x = sin($$\theta$$)

But from where I see it, and I guess that is the problem, if the integral is just provided in the form it is above, then x $$\in$$$$\Re$$

Then why can we substitute x by the function g($$\theta$$) = sin ($$\theta$$) when
g($$\theta$$) $$\in$$ [-1 ; 1]. From my point of view there is a problem when the function we substitute by only gives and small fraction of the possible inputs x can have...

Anyone follow my thoughts? Someone can explain it?

regards,
Rune
Engineering Student

Well to me this technique never seemed anything more than a trick either.

However, it makes a hell of a lot of sense if you think of a triangle:

[PLAIN]http://img401.imageshack.us/img401/3499/triangleb.jpg

$$\int_{a}^{b}\frac{1}{\sqrt{1 \ - \ x^2}} \,dx$$

$$\int_{a}^{b} \ csc( \theta ) \ dx$$

Now, looking at the picture we see that:

$$cos( \theta ) \ = \ \frac{x}{1}$$

$$cos( \theta ) \ = \ x$$

$$sin( \theta ) d \theta \ = \ dx$$

$$\int_{a}^{b} \ csc( \theta ) \ dx$$

$$\int_{a}^{b} \ csc( \theta ) \ sin( \theta ) d \theta$$

$$\int_{a}^{b} \ \frac{1}{sin( \theta )} \cdot \ sin( \theta ) d \theta$$

$$\int_{a}^{b} \ \frac{sin( \theta )}{sin( \theta )} d \theta$$

$$\int_{a}^{b} \ 1 d \theta$$

$$\int_{a}^{b} \ d \theta$$

Remember that when you take the integral you have to get it out of trigonometric form
and into the standard x's etc... you started with. It will involve arctangents or whatever
comes up, you know what I mean...

By the way, this method is extremely helpful in most "fraction-like" integrals, you can
play with your function, i.e. using completing the square under a radical sign or something,
to get really complex looking things to be very simple. Enjoy!

Last edited by a moderator:

You main objection "From my point of view there is a problem when the function we substitute by only gives and small fraction of the possible inputs x can have..." simply isn't correct. It is NOT the case that x can be any real number.

In that particular problem,
$$\int \frac{1}{\sqrt{1- x^2}}dx$$
x can't be "any real number". Since $1- x^2$ is inside a square root, that is possible only if $1- x^2\ge 0$ which is the same as saying that $x^2< 1$ and that immediately leads to $-1\le x\le 1$.

We use such trig substitutions to get rid of square roots- and those square roots force the x values to be appropriate for the trig substitution.

For example, if an integrand involved $\sqrt{a^2- x^2}$, I would use the substitution $x= a sin(\theta)$ so that $\sqrt{a^2- x^2}= \sqrt{a^2- a^2sin^2(\theta)}$$= |a|\sqrt{1- sin^2(\theta)}= |a|cos(\theta)$.

Since $sin(\theta)$ must be between -1 and 1, $x= a sin(\theta)$ must be between -a and a: exactly what is required for $a^2- x^2\ge 0$ so that $\sqrt{a^2- x^2}$ exists!

Last edited by a moderator:

Thank you sponsoredwalk. That was a good explanation! :)

And HallsofIvy, that is my mistake. I am completely gone, hehe. Didn't see that one. Thanks. Now I get it :)

Well to me this technique never seemed anything more than a trick either.

However, it makes a hell of a lot of sense if you think of a triangle:

[PLAIN]http://img401.imageshack.us/img401/3499/triangleb.jpg

$$\int_{a}^{b}\frac{1}{\sqrt{1 \ - \ x^2}} \,dx$$

$$\int_{a}^{b} \ csc( \theta ) \ dx$$

Now, looking at the picture we see that:

$$cos( \theta ) \ = \ \frac{x}{1}$$

$$cos( \theta ) \ = \ x$$

$$sin( \theta ) d \theta \ = \ dx$$

$$\int_{a}^{b} \ csc( \theta ) \ dx$$

$$\int_{a}^{b} \ csc( \theta ) \ sin( \theta ) d \theta$$

$$\int_{a}^{b} \ \frac{1}{sin( \theta )} \cdot \ sin( \theta ) d \theta$$

$$\int_{a}^{b} \ \frac{sin( \theta )}{sin( \theta )} d \theta$$

$$\int_{a}^{b} \ 1 d \theta$$

$$\int_{a}^{b} \ d \theta$$

Remember that when you take the integral you have to get it out of trigonometric form
and into the standard x's etc... you started with. It will involve arctangents or whatever
comes up, you know what I mean...

By the way, this method is extremely helpful in most "fraction-like" integrals, you can
play with your function, i.e. using completing the square under a radical sign or something,
to get really complex looking things to be very simple. Enjoy!

I became very interested in sponsoredwalk's method of doing the substitution.
It is more direct than using x = sin theta to simplify 1 - x^2, and appears quite versatile.
BUT
I found that for some integrals the result is wrong.
Then I noticed that EVEN sponsoredwalk's example appears incorrect to me !

Look and see that the integral works out to be "theta"
The triangle shows cos theta = x so arccos x = theta.
But the correct antiderivative is arcsin x NOT arccos x.

Can someone explain why the method fails and the result incorrect ?

Note that if one changes the triangle so sin theta = x
it comes out right.

Thanks for any inputs or comments

Last edited by a moderator:

I'm sorry, if you look at my original derivation I set x = cosθ. When I took the derivative of
cosθ I forgot to include the minus sign when writing in LaTeX. If you include the minus sign
then you get the correct answer.

$$- \ \int \ d \theta \ = \ - \ \theta \ + \ C$$

If you look at the picture we can see that: $$\cos \ \theta \ = \ x$$$$\ \theta \ = \ \cos^{-1} \ ( \ x \ ) \ = \ arccos \ ( \ x \ )$$

So plugging this all in we get:

$$- \ \int \ d \theta \ = \ - \ \theta \ = \ - \ arccos \ ( \ x \ ) \ + \ C$$

Still not convinced?

$$y \ = \ - \ arccos \ ( \ x \ )$$

$$cos \ y \ = \ - \ x \$$

$$- \ \sin \ y \ \frac{dy}{dx} \ = \ - \ 1$$

$$- \ \frac{dy}{dx} \ = \ - \ \frac{1}{ \sin \ y }$$

Getting rid of the minus signs!

$$\frac{dy}{dx} \ = \ \ \frac{1}{ \sin \ y }$$

Now, from $$\sin^2 \ y \ + \ \cos^2 \ y \ = \ 1$$ we solve for sin: $$\sin \ y \ = \ \sqrt{1 \ - \ \cos^2 \ y}$$

$$\frac{dy}{dy} \ = \ \frac{1}{ \sqrt{1 \ - \ \cos^2 \ y} }$$

Now, when we were taking the derivative in the first place we noticed that x = cos y
so obviously cos²y = x² so we get:

$$\frac{dy}{dx} \ = \ \frac{1}{ \sqrt{1 \ - \ x^2} }$$

Which was the original integral we began with.

So plugging this all in we get:

$$- \ \int \ d \theta \ = \ - \ \theta \ = \ - \ arccos \ ( \ x \ ) \ + \ C$$

Still not convinced?

Wait a minute. Not so fast :-)

"in this manner" is = - arccos x.

And we know from standard tables and the classic method of
substituting sin theta for x, the correct result is arcsin x.

So are you saying that arcsin x is NOT the correct result ?
Or do you contend that -arccos x = arcsin x ?

What you did is verify the result by differentiating, thus invoking the
Fundamental Rule of Calculus ... Integration and Differentiation are
inverse functions. But that is irrelevant. The method produces the wrong
answer depending on how one assigns the sides of the right triangle.

Are we having fun yet ?

Haha, forget the tables Follow the logic, I'll set everything up again.

$$\int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx$$

Looking at the picture I drew you'll see that we can rewrite the integral as

$$\int \ csc \ \theta \ dx$$

Now, again if we look at the picture we can express $$\cos \ \theta \ = \frac{x}{1}$$

This is the same as $$\cos \ \theta \ = \ x$$. If I take the derivative I get:

$$- \ \sin \ \theta \ d \ \theta \ = \ dx$$

I can rewrite the integral as:

$$\int \ csc \ \theta \ dx \ = \ - \ \int \ csc \ \theta \ \sin \ \theta \ d \ \theta$$

We know that $$\csc \ \theta \ = \ \frac{1}{ \sin \ \theta }$$ so

$$- \ \int \ csc \ \theta \ \sin \ \theta \ d \ \theta \ = \ - \ \int \frac{1}{ \sin \ \theta } \ \sin \ \theta \ d \ \theta \ = \ - \ \int \frac{ \sin \ \theta }{ \sin \ \theta } \ d \ \theta \ = \ - \ \int \ d \ \theta$$

The final integral is:

$$- \ \int \ d \theta \ = \ - \ \theta \ + \ C$$

Remember that $$\cos \ \theta \ = \ x$$. I can then take the inverse to
re-express the angle in terms of x's.

$$\cos \ \theta \ = \ x$$

$$\theta \ = \ \arccos \ ( \ x \ )$$

$$- \ \int \ d \theta \ = \ - \ \theta \ = \ - \ arccos \ ( \ x \ ) \ + \ C$$

Actually, you're going to have to forgive me, at the very end of the last answer I made a mistake.
When I wrote "x = cos y so obviously cos²y = x²" I forgot
that I actually set x = - cos y. When I take the derivative of inverse functions I usually go to the side
of the page and rederive them from basics rather than memorizing them. I should never have
included the minus when I was deriving the original integral from - arccos x. It kind of invalidated
everything I wrote in the last response from "Still not convinced?" onwards

One last time!

Still not convinced? I can use differentiation to check whether I've gotten the right answer
seeing as integration and differentiation are inverse processes. I should be able to, by
differentiating, end up with the exact same integral as I started with.

$$y \ = \ arccos \ ( \ x \ )$$

$$cos \ y \ = \ x \$$

$$- \ \sin \ y \ \frac{dy}{dx} \ = \ 1$$

$$- \ \frac{dy}{dx} \ = \ \frac{1}{ \sin \ y }$$

$$\frac{dy}{dx} \ = \ - \ \frac{1}{ \sin \ y }$$

Now, from $$\sin^2 \ y \ + \ \cos^2 \ y \ = \ 1$$ we solve for sin: $$\sin \ y \ = \ \sqrt{1 \ - \ \cos^2 \ y}$$

$$\frac{dy}{dy} \ = \ \frac{1}{ \sqrt{1 \ - \ \cos^2 \ y} }$$

Now, when we were taking the derivative in the first place we noticed that x = cos y
so obviously cos²y = x² so we get:

$$\frac{dy}{dx} \ = \ - \ \frac{1}{ \sqrt{1 \ - \ x^2} }$$

This is the derivative of arccosx, if we want the derivative of minus arccos x we just
negate the sign of this final result and we get

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

Which was the original integral.

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Haha, forget the tables Follow the logic

This is the derivative of arccosx, if we want the derivative of minus arccos x we just
negate the sign of this final result and we get

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

Which was the original integral.

solution to be theta = - arccos x. Your step by step solution looks correct.

But the problem is that this contradicts
1/ All integration tables that show the integral to be arcsin x
2/ The solution you get if you interchange the non hypotenuse sides
and solve with sin theta = x
3/ The solution one obtains if using the traditional substitution of
x = sin theta.

So I repeat my questions ...
Are you saying that arcsin x is NOT the correct result ?
Or do you contend that -arccos x = arcsin x ?

I do not think there is any interval for a definite integration of this
integrand which will give the same numerical result for both solutions.

This is an interesting problem.

Last edited:

Testing

$$\int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx$$

Sorry for the late response. Here is a way to test what I've done for yourself.
Take the definite integral of the function from -0.5 to 0.5 using this triangle method
I've used but do two integrals. First do the integral the way I did it and evaluate it.
Then do the integral changing the legs, i.e. set $$\sqrt{ 1 \ - \ x^2}$$ to be
the bottom leg. Let me know.

I think that this discrepancy between the two solutions is accounted for by having different constants of integration.

If you do the two definite integrals you get the same answer.

If you do the integral my way you get $$\int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx \ = \ - \ cos^{-1}(x) \ + \ C$$

If you do the integral with the $$\sqrt{1 \ - \ x^2$$ on the other leg

you get $$\int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx \ = \ sin^{-1}(x) \ + \ C$$

There is a question of signs as well as the constant of integration at play.
If you follow the logic accurately you'll get the right answer & you always have
differentiation to test whether you've gotten the right answer or not.
For those integrals you can't easily form a right triangle just use some algebra
like completing the square or factoring etc... to get what you need

Well now I can see that the derivatives of sine x and cosine x differ only in sign.
So that means the integrals of this integrand also differ only in sign.
So arcsin x does indeed = minus arccos x.

What is interesting is why interchanging the right triangle sides produces the sign change.

Note that the exact same thing occurs with arctan x and arccot x
AND arcsec x and arccsc x. Their derivatives are identical except for sign.

Why do I love having my brain twisted by Mathematics ?
I hope its not a masochistic desire ... hahahahaha

Cheers

## 1. What is trigonometric substitution and why is it used?

Trigonometric substitution is a mathematical technique used to simplify integrals involving trigonometric functions. It is used because it allows for the integration of more complicated expressions by substituting them with simpler trigonometric functions.

## 2. How does trigonometric substitution work?

Trigonometric substitution involves replacing a variable in an integral with a trigonometric function. This allows for the integration of more complex functions by using trigonometric identities and the properties of trigonometric functions.

## 3. When should I use trigonometric substitution?

Trigonometric substitution is typically used when the integral involves square roots of quadratic expressions or expressions involving the sum or difference of squares. It is also useful for integrals involving rational functions with trigonometric expressions in the numerator or denominator.

## 4. Are there any specific guidelines for using trigonometric substitution?

Yes, there are a few guidelines to keep in mind when using trigonometric substitution. It is important to choose the correct trigonometric substitution by identifying the appropriate trigonometric identity to use. It is also important to check for any restrictions on the domain of the integral, and to make sure the resulting integral can be easily evaluated.

## 5. Can trigonometric substitution be used for all integrals?

No, trigonometric substitution can only be used for specific types of integrals. It is not a universal technique and may not work for all integrals. In some cases, other integration techniques such as u-substitution or integration by parts may be more suitable.

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