- Introduction to trigonometric substitution
- Substitution with x=sin(theta)
- More trig sub practice
- Trig and u substitution together (part 1)
- Trig and u substitution together (part 2)
- Trig substitution with tangent
- More trig substitution with tangent
- Long trig sub problem
- Trigonometric substitution
Example of using trig substitution to solve an indefinite integral. Created by Sal Khan.
Want to join the conversation?
- Why is the integral of d(theta) = theta?(27 votes)
- All d(theta) means is that you are integrating with respect to theta. It's just like how dx means to integrate with respect to 'x'. An indefinite integral of (1/sqrt)dx would be (x/sqrt). So if we took the integral of (1/sqrt) with respect to theta, the answer is just (1/sqrt)*theta. Don't forget that (1/sqrt) is a constant. To take the antiderivative of any constant, you just tack on the variable that you are dealing with.(53 votes)
- If I ever got a question like this on my exam (which won't be 2 years from now), how would I go about doing it?
How will it strike me to take 3 common, to convert the question into trig identities, etc etc, so many more things?
Would anyone be able to do a question like this if they saw it for the first time in their lives?
(assuming they are not mathletes like Sal of course)(17 votes)
- The biggest clue that you need to use trig substitution that Sal didn't mention is that the term 1/sqr(3-2x^2) looks very similar to the derivative of arcsin(x), which you've learned as 1/sqr(1-x^2). So naturally, it's safe to guess that the antiderivative of 1/sqr(3-2x^2) is an inverse sine function.
Of course, by the time the average calc student figures that out they will probably be slamming their head on their desk trying to do this problem with u-substitution.(35 votes)
- Someone help me out with the final answer! Arcsine has domain restrictions, how is this accounted for? do you need to simplify it further for a more complete answer? does this mean the value which arcsine is being evaluated at needs to fit within the domain, therefore x needs to be within the domain?(16 votes)
- I had the same question and I didn't feel satisfied until I proved it to myself :) Here's the proof I wrote for peter.s.vdm's answer.
- If I substitute sin^2theta = cos^2theta I get a negative sign in the final result. How can two answers be possible or am I making some mistake?(5 votes)
- At2:07, Sal takes (2/3)x^2 = sin^2(theta).
But shouldn't it be (2/3)x^2 = m.sin^2(theta) where m is any arbitrary constant?
The range of the sine function is only from -1 to 1.
In case our x is greater than, say, 3, take 6 for example, then (2/3)x^2 cannot equal sin^2(theta), since sin^2(theta) can never equal 24, since it can equal 1 at max.
This where the constant 'm' will come in handy.
Why is it okay just to use sin(theta) ?(6 votes)
- You are wrong that the sine maps only to the range [-1,1]. While its true if you are dealing with the real numbers, it is not true if you are dealing with complex numbers. For example, sin(pi/2-i ln(2+sqrt(3)))=2(5 votes)
- At2:17, when he takes the square root of the equation, should he not put in a plus or minus (+-), since squaring a positive or a negative will give you the same result?(5 votes)
- When you make the substitution x=a*sin(theta), you are limiting the domain of x from all real numbers to the interval [-a, a]. Wouldn't this mean the answer is only valid for x in the interval [-a, a]? Also, how can we assume that sqrt(sin^2 x) = sin x and not -sin x?(3 votes)
- EXCELLENT QUESTION!
I am pretty sure Sal deals with these nuances of trig sub in other videos.
We can make the trig substitution x = a sin θ provided that it defines a one-to-one function. This can be accomplished by restricting θ to lie in the interval [-π/2, π/2] (for cos and sin).
The point of trig sub is to get rid of a square root, which by its very nature also has a domain restriction. If we change the variable from x to θ by the substitution x = a sin θ, then we can use the the trig identity 1 - sin²θ = cos²θ which allows us to get rid of the square root sign, since:
√(a² - x²) = √(a² - a²sin²θ) = √(a²(1 - sin²θ)) = √(a²cos²θ) = a|cosθ|
And since θ is in [-π/2, π/2], we can say that a|cosθ| = acosθ.(8 votes)
- why is 1/sqrt2*S d theta = 1/sqrt2 theta + C?????(2 votes)
- S d(theta) = is the same thing as = S 1 d(theta) and the antiderivative of one is theta. because your taking the antiderivative with RESPECT to theta.(8 votes)
- how do you solve for theta like sal did in2:46?(2 votes)
- inverse trigonometric functions
Let's say I have the indefinite integral 1 over the square root of 3 minus 2x squared. Of course I have a dx there. So right when I look at that, there's no obvious traditional method of taking this antiderivative. I don't have the derivative of this sitting someplace else in the integral, so I can't do traditional u-substitution. But what I can do is I could say, well, this almost looks like some trig identities that I'm familiar with, so maybe I can substitute with trig functions. So let's see if I can find a trig identity that looks similar to this. Well, our most basic trigonometric identity-- this comes from the unit circle definition-- is that the sine squared of theta plus the cosine squared of theta is equal to 1. And then if we subtract cosine squared of theta from both sides, we get-- or if we subtract sine squared of theta from both sides, we could do either-- we could get cosine squared of theta is equal to 1 minus sine squared of theta. We could do either way. But this, all of a sudden, this thing right here, starts to look a little bit like this. Maybe I can do a little bit of algebraic manipulation to make this look a lot like that. So the first thing, I would like to have a 1 here-- at least, that's how my brain works-- so let's factor out a 3 out of this denominator. So this is the same thing as the integral of 1 over the square root of-- let me factor out a 3 out of this expression. 3 times 1 minus 2/3x squared. I did nothing fancy here. I just factored the 3 out of this expression, that's all I did. But the neat thing now is, this expression looks a lot like that expression. In fact, if I substitute, if I say that this thing right here, this 2/3x squared, if I set it equal to sine squared theta, I will be able to use this identity. So let's do that. Let's set 2/3x squared, let's set that equal to sine squared of theta. So if we take the square root of both sides of this equation, I get the square root of 2 over the square root of 3 times x is equal to the sine of theta. If I want to solve for x, what do I get? And, well, we're going to have to solve for both x and for theta, so let's do it both ways. First, let's solve for theta. If we solve for theta, you get that theta is equal to the arcsine, or the inverse sine, of square root of 2 over square root of 3x. That's if you solve for theta. Now, if you solve for x, you just multiply both sides of this equation times the inverse of this and you get x is equal to-- divide both sides of the equation by this or multiply it by the inverse-- is equal to the square root of 3 over the square root of 2 times the sine of theta. And we were going to substitute this with sine squared of theta, but we can't leave this dx out there. We have to take the integral with respect to d theta. So what's dx with respect to d theta? So the derivative of x with respect to theta is equal to square root of 3 over square root of 2. Derivative of this with respect to theta is just cosine of theta, and if we want to write this in terms of dx, we could just write that dx is equal to square root of 3 over the square root of 2 cosine of theta d theta. Now we're ready to substitute. So we can rewrite this expression up here-- I'll do it in this reddish color-- I was using that, let me do it in the blue color. We can rewrite this expression up here now. It's an indefinite integral of-- dx is on the numerator, right? Instead of writing this 1 times dx, I could have just written a dx up here. That could be a dx just like that. You're just multiplying it times dx. So what's dx? dx is this business. I'll do it in yellow. dx is this right here. So it's the square root of 3 over the square root of 2 cosine theta d theta. That's what dx was. Now, the denominator in my equation, I have the square root of 3 times-- now it's 1 minus. Now I said 2/3x squared is equal to sine squared of theta. Now how can I simplify this? Well, what's 1 minus sine squared of theta? That's cosine squared of theta. So this thing right here is cosine squared of theta. So my indefinite integral becomes the square root of 3 over the square root of 2 cosine theta d theta, all of that over the square root of 3 times the cosine squared of theta. That just became cosine squared of theta. So let's just take the square root of this bottom part. So this is going to be equal to-- I'll do an arbitrary change of colors-- square root of 3 over the square root of 2 cosine of theta d theta, all of that over-- what's the square root of this? It's equal to the square root of 3 times the square root of cosine squared, so times cosine of theta. Now, this simplifies things a good bit. I have a cosine of theta divided by a cosine of theta, those cancel out, so we'll just get 1, and then I have a square root of 3 up here divided by a square root of 3, so those two guys are going to cancel out, so my integral simplifies nicely to 1 over square root of 2 d theta. Or even better, I could write this-- this is just a constant term, I could take it out of my integral-- it equals 1 over the square root of 2 times my integral of just d theta. And this is super easy. This is equal to 1 over the square root of 2 times theta plus c. Plus some constant. I mean, you could say that the integral of this is theta plus c and then you'd multiply the constant times this, but it's still going to be some arbitrary constant. I think you know how to take the antiderivative of this. But are we done? Well, no. We want to know our indefinite integral in terms of x. So now we have to reverse substitute. So what is theta? We figured that out here. theta is equal to arcsine the square root of 2 over the square root of 3x. So our original indefinite integral, which was all of this silliness up here, now that I reverse substitute for theta or put x back in there, it's 1 over the square root of 2 times theta. theta is just this, is just arcsine of square root of 2 over square root of 3 x, and then I have this constant out here, plus c. So this right here is the antiderivative of 1 over the square root of 3 minus 2x squared. So hopefully you found that helpful. I'm going to do a couple of more videos where we go through a bunch of these examples, just so that you get familiar with them.