- 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
All of the substitution! Created by Sal Khan.
Want to join the conversation?
- I don't understand where sal gets x=asin(theta) from a^2-b^2.... PLEASE HELP(13 votes)
- Sal explains this in the first video, Intro to trig substitution, but it's not intuitive to me either so it's worth going over again:
If you have a right triangle with hypotenuse of length a and one side of length x, then:
x^2 + y^2 = a^2 <- Pythagorean theorem
where x is one side of the right triangle, y is the other side, and a is the hypotenuse.
So anytime you have an expression in the form a^2 - x^2, you should think of trig substitution. Now here's where the trig comes in:
Drawing our right triangle:
sin theta = opposite / hypotenuse
sin theta = x/a
solve for x:
a * sin theta = x(41 votes)
- cant i take
is it correct.. cuz i'm gettin a different answer if i do that!!(3 votes)
- ok this is from 8 years ago, but I'll answer anyways lol.
the derivative of your u isn't 3cos^2(x)dx, it's 3sin^2(x)cos(x)
(I'm using x instead of theta)(3 votes)
- At7:50, when we are taking the integral of the function, shouldn't the result be also multiplied by "u", as the integral of "du" is just "u+C"?
Thank you!(2 votes)
- Why did he substitute u=cos(theta) instead of u=sin(theta)?(2 votes)
- Sal created an extra sin(theta) so he could use it as du. Sin(theta) is the derivative of -cos(theta). Therefore u must be cos(theta).(2 votes)
- it will work with u-substitution directly with u = sqrt ( 9 - x^2 )(2 votes)
- if u= sqrt (9-x^2), then how does one account for du = 2x without an x in the numerator?(2 votes)
- Sal: You're not getting very far with u-substitution.
Me: You're definitely not trying hard enough! With proper u-substitution, you can solve this problem in a smaller number of steps and do fewer mistakes. Substitute
The integral thus becomes,
u^2 = 9 - x^2
x du = -u du
How is this not easier to solve? :D
∫u^4 - 9u^2 du(2 votes)
- why couldnt we assign cos theta to x?(1 vote)
- Cos theta would work just as well, and the choice of which one to use is fairly arbitrary. There seems to be a general preference for sin, maybe to avoid introducing a negative sign in dx (derivative of sin is cos, but derivative of cos is -sin). That wouldn't be a problem, just a place where you could make a mistake if you aren't careful.(2 votes)
- Where i can find the proof, that if i have a^2 - x^2.
i can substitu x as equal to, asin(theta) ?(1 vote)
- You can see the introductory video for trigonometric substitution, while not a proof, it shows why that is a valid substitution: https://www.khanacademy.org/math/integral-calculus/integration-techniques/trig_substitution/v/introduction-to-trigonometric-substitution(2 votes)
Let's attempt to take the antiderivative, or the indefinite integral, of x to the third times the square root of 9 minus x squared dx. And you might attempt to do something like use substitution, but you'll find that you're not getting very far. And we have a big clue here. We have something of the form-- so 9 minus x squared, that can be viewed as the same thing as 3 squared minus x squared. And any time you have the form a squared minus x squared, it might be useful to make the substitution that x is equal to a sine theta. Now, why would that be? Well, then a squared minus x squared would become a squared minus a squared sine squared theta, which is the same thing as a squared times 1 minus sine squared theta. And I think you see where this is going. This is a squared times cosine squared of theta, which might be a useful simplification. So let's do it over here. In this case, our a is 3. So let's make the substitution that x is going to be equal to a sine theta, or three times sine of theta. And then we're going to also have to figure out what dx is equal to. So if you take the derivative, we will get dx. We could have dx d theta is equal to 3 cosine theta. Or if we wanted to write it in differential form, we could write that dx is equal to 3 times cosine theta d theta. This is just the derivative of this with respect to theta. And we're ready to substitute back. Our original expression now becomes-- I'll write it in that original green-- 3 sine theta to the third power, which is the same thing as 27 sine-- actually, let me color code it just so you know what parts I'm doing. So this part right over here, x to the third, is now going to become 27 sine to the third power of theta, or sine theta to the third power. And then all of this business, this is going to be the square root of 9 minus x squared, so minus 9 sine squared theta. And then dx-- let me do this in a new color-- dx right over here is going to be equal to-- that's not a new color. dx is going to be equal to all of this business, so times 3 cosine theta d theta. And now let's see if we can simplify this business a little bit. Let me do this over to the side. This thing right over here can be written as 9 times 1 minus sine squared theta, which is equal to the square root of 9 times cosine squared theta. And we can assume that cosine theta is positive, as we did in the last video. And so this is going to be equal to 3 cosine theta in orange. So this right over here is 3 cosine theta. And so what does this simplify to? We have a 27 times 3 is 81 times 3 is going to be 243. So this is going to be 243-- I'll put it out front-- times the integral of-- let's see, we're going to have sine cubed theta. And then you're going to have cosine theta times cosine theta, or we could say cosine squared theta. That's this term right over here and this term right over there, and of course, d theta. I think I've taken care of everything. So it might not look like I've simplified it a lot, because, hey, look, this still doesn't seem like a trivially easy problem to solve, but we are getting closer. Now this turns into just a classic u substitution problem. And it's not obvious just yet. There's actually a little bit of a layer of a technique to figure out first. How do you do u substitution right over here? And the key when you have powers of trig functions, especially when you have one of them as an odd power, what you want to do is separate one of those odd powers out so you can kind of construct a u substitution problem. So let's do that. So this is going to be equal to 243 times the integral of-- sine cubed theta I can rewrite as sine of theta-- actually, let me write it this way-- as sine squared theta cosine squared theta. And then I still have 1 sine theta right over here, d theta. And what I'm trying to do is turn this expression into something where I can do u substitution, and as you could imagine, maybe where the du has to deal with sine theta d theta. And it would if I can get my u being equal to cosine theta. Then my du is going to be negative sine theta d theta. So let me see if I can do that. So if I say that sine squared theta is the same thing as 1 minus cosine squared theta, then this whole thing becomes 243 times the integral of 1 minus cosine squared theta times cosine squared theta times sine theta d theta. And I want to be very clear what I did over here. So we used some trig substitution to get to this point right over here. And at this point, I took one of the sine thetas out-- I separated it right over here-- and then I converted this to an expression in terms of cosine theta. Now, the whole reason why I did this is right over here I have a function of cosine theta, and then I have something that's pretty close to the derivative of cosine theta right over here. So this is now ripe for u substitution. So let's do u substitution. If I have a function of something and then I have this derivative, maybe u should be equal to that something. So let me set u as being equal to cosine of theta. Then du is going to be equal to negative sine of theta d theta. Well, I have a sine theta d theta. I can multiply that times a negative as long as I put a negative out here. I'm multiplying by a negative twice. I'm not changing the value. So notice, now this right over here is du, and this right over here is a function of u. So let's write it that way. All of this business is going to be equal to negative 243 times the integral of 1 minus u squared times u squared. And then this right over here is just our du. So this is just du. Now, this is pretty straightforward. We can just multiply our u out, and this will become negative 243. So this is all equal to negative 243 times the indefinite integral of u squared minus u to the fourth-- I'm just distributing the u squared-- du. Now, this is pretty straightforward to take the antiderivative of. This is negative 243 times the antiderivative of u squared is u to the third over 3. Antiderivative of u to the fourth is u to the fifth over 5. And of course, we're going to have a plus c out here. And just so that we can get rid of this negative, we can swap this. We can distribute the negative sign. So this one becomes negative. That one becomes positive. And we get all of this business as being equal to 243 times u to the fifth over 5 minus u to the third over 3 plus c. And you might say, wow, finally, we are done. But we aren't done. We have everything in terms of u while our original integral was in terms of x. So the next video, we're going to unwind all the substitution. We're going to try to take this expression right over here and write it in terms of x. So we're going to have to go from u to theta to x, because we've done two rounds of substitutions.