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Substitution with x=sin(theta)

When you are integrating something that has the expression (1-x^2), try substituting sin(theta) for x. Created by Sal Khan.

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  • blobby green style avatar for user Aiden McArdle
    How does x=2sin(theta)? I dont understand how Sal got that value
    (43 votes)
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    • leaf green style avatar for user Roma
      From the form a^2 - x^2 he said x could be written as x=asin(theta)
      Then you got that 8 - 2x^2 could be rewritten as 2(2^2 - x^2). There you have your "a" on the thing you got now inside the parenthesis, and it is equal to 2.
      So if you replace it on what you had as x=asin(theta) you get x=2sin(theta)
      (18 votes)
  • piceratops ultimate style avatar for user Ali Taqi
    why is a^2-x^2 = a*sin(theta)
    (14 votes)
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  • leaf green style avatar for user Eirik Søreng
    Hi, I got a quite simular integral infront of me atm. Can anyone help me?
    It is as follows: Integrate (1/(a^2+x^2))dx
    (7 votes)
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    • piceratops tree style avatar for user David
      Try to think about the pythagorean identities associated with trigonometric functions when you see a problem like this. You want to substitute a function in there, so we choose tan(theta) since it is related to sec(theta) by tan^2(theta) + 1 = sec^2(theta).
      So, in order for this substitution to work out okay, you're letting x=a*tan(theta) so that when you write it out, you will end up with a^2+(a*tan(theta))^2 in your denominator. Simplifying leads to a^2+(a^2 * tan^2(theta)), and factoring the a^2 out gets: a^2(1+tan^2(theta)). Much like this video, it is basically the same process, just keeping in mind our relationships to tangent and secant instead of sine and cosine.
      That said, ta-da! You have the definition for sec^2(theta) in your denominator now, right? But we have to make sure to convert that dx to d(theta). if x = a*tan(theta) then dx = a*sec^2(theta)*d(theta). So, substituting the two parts we figured out gives us (a^2 * sec^2(theta)) in the denominator and sec^2(theta)d(theta) in the numerator. These sec^2(theta)s can cancel leaving 1/(a^2) d(theta), and since (a) is a constant, your integral simplifies to 1/(a^2) * (theta) + C. All that's left to do is convert back to the original terms. As Sal mentions around , we've already defined theta. Since x = a*tan(theta), tan(theta) = (x/a), and (theta) = arctan(x/a). So your integral simplifies to 1/(a^2) * arctan(x/a) + c.
      Hope that helps!

      Edit: I just found a link to the wikipedia page for Trig substitution, and it pretty much sums everything up neatly if you want to reference it as you get comfortable with these kind of problems. http://en.wikipedia.org/wiki/Trigonometric_substitution
      (17 votes)
  • aqualine ultimate style avatar for user Shubham
    why is dx= 2cosine(theta)?
    (4 votes)
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  • mr pants teal style avatar for user Eli Spina
    What determines how/when you can substitute one thing for another? U-substitution seems to be applicable in almost any situation because you preserve the original function and back-substitute. So, why isn't trig substitution a transformation of the original function?
    (4 votes)
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    • piceratops ultimate style avatar for user Connor Hatch
      But you are "back-substituting" in trig substitution as well
      Trig substitution just seems to be a spin on U-Substitution
      When we first make our substitution in this problem we are saying that:
      x = 2sin(theta)
      Sal later goes on to clarify that:
      (theta) = arcsin(x/2)
      This is still in terms of the x we originally started off with
      Finally, at the very end of this integration, we "back-substitute" arcsin(x/2) for theta, this is the "back-substitution" that you are looking for like in U-Substitution. And because our back-substitution is in terms of the original x, we are preserving the original function
      (5 votes)
  • leaf green style avatar for user TeraVolt
    Can anyone please integrate 1/root(1-x^3)? I want to see the method that is required to solve this integral. I don't see how it is possible to use trig substitution for this problem.
    (3 votes)
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  • leaf grey style avatar for user Musab Rahman
    Couldn't we just approach it like we did in the previous video. We would take 2√2 as the hypotenuse, and √2x as the opposite side for angle 'theta' and solve. The domain doesn't break.
    (3 votes)
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  • mr pink red style avatar for user Mr. Awesome Pavlovsky
    Why does Sal substitute in x = r sin(theta)? would it not make more sense to substitute x = r * cos(theta) [as that is true per polar coordinates]?
    (3 votes)
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    • leaf blue style avatar for user Stefen
      The problem is in Cartesian system and we are using identities from the Pythagorean Theorem.

      There are problems that benefit (are made easier) by the conversion from Cartesian to Polar systems (see pre calc for the conversions) and Integration Applications for integrating using Polar coordinates.
      (2 votes)
  • male robot donald style avatar for user harry park
    At what point did we actually solve for the anti-derivative.
    This entire video I just feel like we've been substituting in and out and in and out, and then re-writing some stuff, but I never understood the part where we actually solved for the anti-derivative.
    Just a lot of substituting got me confused.
    (1 vote)
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    • leaf grey style avatar for user Nabla ∇
      I think it's much clearer this process if you think about the purpose behind a trig substitution of a square root with the form √(a²±x²) in the denominator or in the numerator.

      We can't integrate an expression like that. The first thing one would try to do, is to see whether the derivative of the inside function, a²±x², is in the integral to try an u-substitution.

      Since it isn't one, a clever thing to attempt is to find a way to right the inside expression ,a²±x², as just one term so that the square root becomes something that we can get rid of.

      The neat thing is that √(a²±x²) resembles something that we know of: a relation between the two sides of a right triangle that gives us the remaining (one) side. It may be the hypotenuse (√(a²+x²)) or one of the two sides (√(a²-x²)). We also know that the sides of a right triangle can be written in terms of the cosine and sine functions. That's cool, but by doing so we'll want to have a way that gets us back to x (that's is the original variable of the integral). Here is where arctan and arcsin come into play.

      What Sal does in the video is exactly that: Write what is inside of the square root as one expression in order to get rid of it and then find a way to go back to x. The rest is just part of the integration process we've seen so far.

      Hope this helps.
      (6 votes)
  • leaf orange style avatar for user twilight.imperil
    Can we ever NOT assume that the abs. value is positive? My teacher told us it will always be positive, but I'm not sure if he was just talking about sec(x) and tan(x).
    (1 vote)
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    • piceratops ultimate style avatar for user Just Keith
      If you want to get really nit-picky about it, your teacher is wrong. An absolute value is always non-negative, but not always positive. The only exception is 0, which is neither positive nor negative. Except for 0, the absolute value is always positive.

      So, the correct term is "non-negative" since that includes all positive numbers and 0.
      (6 votes)

Video transcript

Let's see if we can evaluate this indefinite integral. And the clue that trig substitution might be appropriate is what we see right over here in the denominator under the radical. In general, if you see something of the form a squared minus x squared, it tends to be a pretty good idea, not always, but it's a good clue that it might be a good idea to make the substitution x is equal to a sine theta. Because if you do that, then this will become a squared minus a squared sine theta. And if you factor out the a squared, you can start leveraging-- this is a squared-- you can leverage one of the most basic trig identities that this right over here is cosine squared theta and maybe simplify the expression. Now, you're probably saying, well, this 8 minus 2x squared, it's not as obvious that it's a squared minus x squared. But we could simplify this. Or I guess we can write it in a way that it starts to have this pattern. You can rewrite 8 minus 2-- let me write it right under it. You could write 8 minus 2x squared as, if we factor out a 2, as 2 times 4 minus x squared. And now this very clearly has a pattern a squared minus x squared. You could write this as 2 times 2 squared minus x squared. So in this case, a would be equal to 2. So let's make that substitution. Let's make the substitution that x is going to be equal to 2 sine theta and dx is going to be equal to 2 cosine theta d theta. So what's this part under the expression going to be? Well, we already started simplifying it right over here. It's going to become 2 times 2 squared minus x squared. x squared is 2 sine theta, so x squared is going to be 2 squared sine theta squared. And now we can factor out the 2 squared. So this is going to be 2 times 2 squared times 1 minus sine squared theta. 2 times 2 squared, well that's just going to be 8, times cosine squared theta. That's what we have under the radical. So let's do that. Let's rewrite this thing up here. So we're going to have-- and I'll take the pi outside of the integration. So we're going to have pi times dx. dx is 2 cosine theta d theta. So it's going to be-- let me make it clear. So dx-- I want to do that in blue. dx right over there is 2 cosine theta d theta. So let me write that, 2 cosine theta. And I'll write the d theta out here. I could have written it in the numerator. And then here in the denominator, I'm going to the square root of this business, the square root of 8 cosine squared theta. So the square root of that is going to be 2 square roots of 2. The square root of 8 is 2 square roots of 2. So let me write this. So let me make it clear what I'm doing. So this right over here is going to be the square root of this, which is 2 square roots of 2. That's the square root of 8. And the square root of cosine squared theta is going to be cosine theta. Now, you might be saying, hey, wait, if I take the square root of something squared, then wouldn't that just be the absolute value of cosine theta? In order to take away the absolute value, I'd have to assume that cosine theta is positive. But we can make that assumption that cosine theta is positive, because if we look right here at this part of our substitution, if we wanted to solve for theta, you'd divide both sides by 2, and you'd get x of 2 is equal to sine of theta. Or we could say that theta is equal to arcsin of x over 2. Now the arcsin function, as it is traditionally defined, will return theta that is between negative pi over 2 and pi over 2. And in that range, cosine of theta is always going to be positive. So we don't have to write the absolute value. We know cosine theta is positive. So now we can start to simplify. Cosine theta cancels out with cosine theta. This 2 cancels out with this 2. We can bring this square root of 2 outside. And so we are left with pi over this square root of 2 times the indefinite integral of just d theta. And this is just going to be equal to pi over the square root of 2 times theta plus c. And we're almost done. We just have to rewrite this in terms of x. And we already know that theta is equal to arcsin of x over 2. So we can say that this indefinite integral, or the antiderivative of this expression, is going to be pi over the square root of 2 times arcsin of x over 2 plus c. And we're done. Some people like a square root of 2 in the denominator. If you want to remove it, you can multiply this by square root of 2 over square root of 2, and that will simplify it. But right now, I'll just leave the denominator in irrational form. And this right over here is our antiderivative.