Trig Substitution

[Metronome]

Hello, I'm having a disproportionately difficult time learning trig-substitution compared to integration by parts, u-substitution, and partial fractions (every video tutorial seems to use a slightly different process).


Should assigning the legs to the right triangle be thought of as part of the solution process, or is it more so an explanation/justification of why a^2 - x^2 yields x = a * sin(theta), x^2 + a^2 yields x = a * tan(theta), and x^2 - a^2 yields x = a * sec(theta), analogous to how the unit circle explains sin and cos, but isn't explicitly used very often in trig problems since the few important points on it are memorized?

 

[mmm4444bot]

… (every video tutorial seems to use a slightly different process).

Should assigning the legs to the right triangle be thought of as part of the solution process, or is it more so an explanation/justification of why:

a^2 - x^2 yields x = a * sin(theta)

x^2 + a^2 yields x = a * tan(theta)

and

x^2 - a^2 yields x = a * sec(theta) …

Can you post a link to the video that generated your question? :cool:

 

[tkhunny]

Hello, I'm having a disproportionately difficult time learning trig-substitution compared to integration by parts, u-substitution, and partial fractions (every video tutorial seems to use a slightly different process).


Should assigning the legs to the right triangle be thought of as part of the solution process, or is it more so an explanation/justification of why a^2 - x^2 yields x = a * sin(theta), x^2 + a^2 yields x = a * tan(theta), and x^2 - a^2 yields x = a * sec(theta), analogous to how the unit circle explains sin and cos, but isn't explicitly used very often in trig problems since the few important points on it are memorized?

Not real clear how to answer that. The way to look at it is the way that makes sense to you and works consistently.

There are only three kinds:

\(\displaystyle \sin^{2}(x) + \cos^{2}(x) = 1\) leads to \(\displaystyle \cos^{2}(x) = 1 - \sin^{2}(x)\) or more generally this is useful for 1 - [some function]^2

\(\displaystyle \tan^{2}(x) + 1 = \sec^{2}(x)\) or more generally this is useful for [some function]^2 + 1 or 1 + [some function]^2

\(\displaystyle \tan^{2}(x) + 1 = \sec^{2}(x)\) leads to \(\displaystyle \tan^{2}(x) = \sec^{2}(x) - 1\) or more generally this is useful for [some function]^2 - 1

You can multiply by any necessary constant and it may take some algebraic manipulation to get your structure to look like one of these.

There is also a very general trigonometric substitution that helps with far more things than perhaps it seems it should. In my opinion, this is more for existence exercises, but not everyone agrees with me.

Let's see some of your work.

Inverse Trig Functions are a different conversation. This may be more what you are talking about.

 

[Dr.Peterson]

Hello, I'm having a disproportionately difficult time learning trig-substitution compared to integration by parts, u-substitution, and partial fractions (every video tutorial seems to use a slightly different process).

Should assigning the legs to the right triangle be thought of as part of the solution process, or is it more so an explanation/justification of why a^2 - x^2 yields x = a * sin(theta), x^2 + a^2 yields x = a * tan(theta), and x^2 - a^2 yields x = a * sec(theta), analogous to how the unit circle explains sin and cos, but isn't explicitly used very often in trig problems since the few important points on it are memorized?

Yes, there are a variety of ways to carry out the details; this is a downside of looking at too many videos or websites, and a reason to stick with one textbook!

I draw a triangle when I want to remember which substitution to use; but that's partly because I don't do these everyday, and because I remember things much better when I have a reason for them, so I can reconstruct them if I forget. For a course, you can (if you are good at memorizing arbitrary facts) just learn that list of patterns. How you choose what to do is up to you.

Are you okay with actually carrying out the process? That's what's important.

 

[Metronome]

Can you post a link to the video that generated your question? :cool:

Right triangle/hyperbola is the first step:
https://www.youtube.com/watch?v=KHmw0bH3F-o
https://www.khanacademy.org/math/ca.../v/introduction-to-trigonometric-substitution
https://www.youtube.com/watch?v=Nv3C7q88MqA (I think this might be substitution for a parametric equation)

Right triangle is used, but only late in the solution process:
https://www.youtube.com/watch?v=3lC5AuCFK4c
https://www.youtube.com/watch?v=Al6p3L2l9zY

No triangle/hyperbola:
https://www.youtube.com/watch?v=xeOgstlMOSw
https://www.youtube.com/watch?v=E61Tj4R80qw
https://www.youtube.com/watch?v=yaTEmrd5hOU

Let's see some of your work.

Are you okay with actually carrying out the process? That's what's important.

My usual pattern is to watch a few videos until I know the solution process and it makes at least vague intuitive sense, then attack problems (naturally, not understanding the solution process makes it impossible to solve problems). I guess it's just taking me a long time to understand the solution process for this problem type.

 

[Dr.Peterson]

My usual pattern is to watch a few videos until I know the solution process and it makes at least vague intuitive sense, then attack problems (naturally, not understanding the solution process makes it impossible to solve problems). I guess it's just taking me a long time to understand the solution process for this problem type.

It might help if you pick a typical problem on which you feel not quite sure, and show your work and what you are thinking as you do it. Then some of us might be able to share ideas.

 

[mmm4444bot]

… I guess it's just taking me a long time to understand the solution process for this problem type.

I watched four of the videos, and three of them used the same method (sine substitution), even though the integrands were a little different. Of those three, one video explains how the sine substitution arises (at the beginning), one gave the same explanation (but at the end), and one chose not to include the explanation. This is to be expected, watching different instructors, yet the main patterns are the same.

You may be right, about the time it's taking to feel confident about the main theme. You may just need more exposure and practice with variations on the theme, to see the patterns clearly. It'll get better.

To echo Dr.P, post specific questions you have about particular steps (when you feel unsure), and we can help you get around those roadblocks. :cool:

 

[sinx]

Hello, I'm having a disproportionately difficult time learning trig-substitution compared to integration by parts, u-substitution, and partial fractions (every video tutorial seems to use a slightly different process).


Should assigning the legs to the right triangle be thought of as part of the solution process, or is it more so an explanation/justification of why a^2 - x^2 yields x = a * sin(theta), x^2 + a^2 yields x = a * tan(theta), and x^2 - a^2 yields x = a * sec(theta), analogous to how the unit circle explains sin and cos, but isn't explicitly used very often in trig problems since the few important points on it are memorized?

If i read your question correctly; 'assigning values to the legs of a right triangle' is an explanation, analagous to the unit circle. It is not part of the calculus 'solution process'. It is just trig.

on the other hand, A 'trig substitution' is a technique (used in calculus) to put a complicated integrand into a form you can integrate. You first must spot (see) a trig identity in the integrand. Second you substitute a trig function for the variable.

for example; if the integrand is sqrt(1-x²), you can set x=cos(theta);
now; the integrand is sin(theta)*dx.
[this works because of the trig identity; 1-cos²(theta)=sin²(theta)]

change dx to d(theta); find what dx is from x=cos(theta); i.e. dx/d(theta)=sin(theta);
then you have sin²(theta)*d(theta) in the integrand.
use the double angle formula, cos2A=1-1sin²A. (where A is theta)
now you have a form you can integrate, i.e. 1-cos2A.

 

[Deleted member 4993]

If i read your question correctly; 'assigning values to the legs of a right triangle' is an explanation, analagous to the unit circle. It is not part of the calculus 'solution process'. It is just trig.

on the other hand, A 'trig substitution' is a technique (used in calculus) to put a complicated integrand into a form you can integrate. You first must spot (see) a trig identity in the integrand. Second you substitute a trig function for the variable.

for example; if the integrand is sqrt(1-x²), you can set x=sin(theta);
now; the integrand is cos(theta). [a simple integration]
[this works because of the trig identity; 1-sin²(theta)=cos²(theta)]

Are you talking about the following integration:

\(\displaystyle \displaystyle{\int \dfrac{dx}{\sqrt{1-x^2}}}\) ............ if yes - then things get to be a bit different

substitute

x = sin(\(\displaystyle \theta\))

dx = cos(\(\displaystyle \theta\)) d\(\displaystyle \theta\)

\(\displaystyle \displaystyle{\int \dfrac{dx}{\sqrt{1-x^2}}}\)

\(\displaystyle = \ \displaystyle{\int \dfrac{cos(\theta) d\theta}{\sqrt{1- sin^2(\theta)}}}\)

\(\displaystyle = \ \displaystyle{\int d\theta}\)

\(\displaystyle = \ \displaystyle{\theta} + C\)

\(\displaystyle = \ \displaystyle{sin^{-1}(x)} + C\)

 

[mmm4444bot]

… 'assigning values to the legs of a right triangle' is an explanation, analagous to the unit circle. It is not part of the calculus 'solution process'. It is just trig …

Just trig? Heh. Something prompts me to comment beyond simply noting that algebra and trig drive the calculus machine. :cool:

When 'solution process' refers to a specific algorithm itself (eg: listing of integration steps), then I agree. We don't always include scratch-paper steps.

When 'solution process' refers to the process of solving an exercise, then I would say that modeling with right triangles is part of the process.

The Pythagorean identities you mention are a great way to help recognize which trig substitution is needed for certain integrands. A right triangle model can help recall these identities correctly. I still sketch right triangles and label them, to help me with stuff that I ought to remember but don't.

… on the other hand, A 'trig substitution' is a technique (used in calculus) to put a complicated integrand into a form you can integrate.

for example; if the integrand is sqrt(1-x²), you can set x=cos(theta);
now; the integrand is sin(theta)*dx.
[this works because of the trig identity; 1-cos²(theta)=sin²(theta)]

change dx to d(theta); find what dx is from x=cos(theta); i.e. dx/d(theta)=sin(theta);
then you have sin²(theta)*d(theta) in the integrand.
use the double angle formula, cos2A=1-1sin²A. (where A is theta)
now you have a form you can integrate, i.e. 1-cos2A.

I like your explanation of 'trig substitution' as one of the techniques used to change a difficult integral into a form that you know how to work with. However, the example above seems to need some proofreading. Terminology technicalities aside (eg: notations like dx and dθ are not part of the integrand), there seem to be sign, substitution, or maybe typo errors, in the explanation for integrating sqrt(1-x^2) with respect to x. :cool:

 

[sinx]

Just trig? Heh. Something prompts me to comment beyond simply noting that algebra, trig, precalculus comprise the mechanics that drive the calculus machine. :cool:

When 'solution process' refers to a specific algorithm itself (eg: listing of integration steps), then I agree. We don't always include scratch-paper steps; some people reach a point where they don't even need paper.

Were 'solution process' to denote the overall process of solving an exercise or grasping a big picture, then I would say that modeling with a right triangle (on scratch paper or not) is a part of the process. It sure is for me because I tend to loose instant recall of things, especially things that I can derive. The Pythagorean identities you mention are a great way to help recognize which trig substitution is needed for certain integrands. A right triangle model can assist in correctly recalling these identities.

As an aside, I remember being away from trig for months at a time, from 9th grade through college, and beyond. Trig would suddenly reappear (sometimes like a locomotive on a freight train), and I regularly avoided panic by picturing x- and y-values of points on the unit circle, while recalling sine for vertical sides and cosine for horizontal sides of reference triangles, all to help me not mix up numerical trig values for the special angles. Or, simply using the right-triangle trig definitions of sine and cosine to label a trianglular model or to derive a tangent expression. In these examples, assigning trig expressions to the legs of a right triangle was part of my 'solution process', in a variety of different courses. Long story short, the more ways a student has to visualize models and interpret patterns, the more effective their solution processes become.

I like your explanation of 'trig substitution' as one of the techniques used to change a difficult integral into a form that you know how to work with. However, the example above seems to need some proofreading. Terminology technicalities aside (eg: notations like dx and dθ are not part of the integrand), there seem to be sign, substitution, or maybe typo errors, in the explanation for integrating sqrt(1-x^2) with respect to x. :cool:

Just trig? Heh. Something prompts me to comment beyond simply noting that algebra, trig, precalculus comprise the mechanics that drive the calculus machine.

I was defining calculus as derivative and integral, seperate from trig.
when i said the rt triangle is not part of the "solution process", i meant it is not part of the "calculus solution process", i.e it is not part of derivative or integration.
derivative and integral of trig functions themselves of course is a different matter.

there seem to be sign, substitution, or maybe typo errors, in the explanation for integrating sqrt(1-x^2) with respect to x.

there is at least a sign error; d/d(theta) cos(theta)=-sin(theta); but when you factor out a -1, you have the same integrand.
and I also omitted dx and d(theta).
it would have been better had a wrote out the entire equation; i.e. dy=int(1-x²)dx;
then focused on 1-x²​.

 

[mmm4444bot]

…when i said the rt triangle is not part of the "solution process", i meant it is not part of the … integration.

We agree.

 

[Metronome]

You guys were right. The process isn't actually bad, it was just the triangle part throwing me off.