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## Integral Calculus (2017 edition)

### Course: Integral Calculus (2017 edition)>Unit 6

Lesson 6: Trigonometric substitution

# 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.

## Want to join the conversation?

• 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) • 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)? • 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. • 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]? • 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) • What does arcsin means ?
I mean what does it represents ? • 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) • 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.