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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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• at
How does x=2sin(theta)? I dont understand how Sal got that value
• 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)
• why is a^2-x^2 = a*sin(theta)
• 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)`
• 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
• 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
• why is dx= 2cosine(theta)?
• 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).
• 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?
• 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
• 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.
• 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.
• 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]?
• 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.
• 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.