- 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
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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How does x=2sin(theta)? I dont understand how Sal got that value(41 votes)
- 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)(17 votes)
- why is a^2-x^2 = a*sin(theta)(14 votes)
- What Sal's doing is substituting
x = a∙sin(θ), that is why he used the symbol
⇒and not an equal sign. So the equality becomes:
a²-x² = a² - a²sin²(θ)(9 votes)
- 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)
- 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))^2in 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 around3:12, 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)
- why is dx= 2cosine(theta)?(4 votes)
- Because we defined x = 2sin(theta) as part of the substitution, dx or "the derivative of x" is the derivative of 2sin(theta) or 2cos(theta).(6 votes)
- 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)
- 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)
- 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)
- I looked that integral up in a table. There is a way to solve it, but it is well-beyond the level of a second year calculus course. You would need to use an Elliptic Integral of the First Kind to solve it.(5 votes)
- 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)
- 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)
- 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)
- 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)
- What does arcsin means ?
I mean what does it represents ?(2 votes)
- arcsin is different notation for sin^-1. Its the inverse function of sinx. It outputs the sin value that would equal whatever u input into the arcsin. for example, you know that sin 45=1/(2)^1/2. sin inverse, or arcsin (1/2) would therefore equal 45(4 votes)
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.