If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Introduction to trigonometric substitution

## Video transcript

let's say that we want to evaluate this indefinite integral right over here and you immediately say hey you got the square root of 4 minus x squared in the denominator you could try you substitution but it really doesn't simplify this in any reasonable way how do you tackle this and an insight that you might have is well distinct the square root of 4 minus x squared this looks like something that I might get if I'm dealing with the Pythagorean theorem if I'm solving for non hypotenuse sides of the Pythagorean theorem especially if the high pot if the length of the hypotenuse is 2 this would be 2 squared and then the other side is X then 2 squared minus x squared this would be the length of the other side well let's just run with that then so if we have if we have a and if and if any of you didn't feel that inside immediately it's okay because to be honest I never actually did not feel that inside probably the first time that I saw things like this so let's say you have a right triangle so let's just visualize that insight that we talked about so if the hypotenuse of the right triangle is of length 2 the hypotenuse is of length 2 hypotenuse is of length 2 and let's say this side right over here actually let me do this side let's say this side actually I'm gonna go with this side this side right over here is of length X this side over here is of length X it's a little counterintuitive because normally we associate this side over here with Y but let's just go with it this side right over here is of length X then what would this side right over here be what would this other non hypotenuse length be well just if you solve for it using the Pythagorean theorem you get that it is going to be the square root of the hypotenuse squared 2 squared which is just 4 minus the other side squared so minus x squared well that's interesting that is this expression that is that inside the intuition that we may have had when we saw this right over here but still how does that help us well this is where the trigonometry comes in because if we define if this angle right over here we say this is Theta then what is sine and cosine of theta going to be in terms of these sides well let's see the sine of theta sine of theta is equal to the opposite over the hypotenuse is equal to x over 2 or if you want to solve for x we get X is equal to 2 sine theta well that's interesting what about the cosine of theta cosine of theta is equal to the adjacent side square root of 4 minus x squared over the hypotenuse or if you want to solve for this side it's going to be we could say that the square root of 4 minus x squared is going to be equal to the hypotenuse times the cosine of theta so that's interesting if X is equal to 2 sine theta then this other side then this entire expression simplifies to 2 cosine theta and that seems pretty interesting now so let's make the substitution let's say that X is equal to 2 sine theta and if X is equal to 2 sine theta then DX is going to be equal to 2 cosine theta D theta and then if we if X is equal to 2 sine theta then what is this thing right over here well we just figure it out this thing is 2 cosine theta this thing is equal to 2 so let me do it in that orange color 2 cosine theta this is equal to 2 cosine theta and we were able to do this drawing this right triangle and using the the right the sohcahtoa definition of these trig functions and obviously we could use unit circle that's kind of an extension of these but you could also do it if you say hey look you know if this is 2 sine theta you could use the Pythagorean identity our trigonometric identities and you would see that this entire expression if you did it that way would simplify to 2 cosine theta but now let's just run forward and let's see if we can evaluate it using this substitution so this is going to be the indefinite integral so DX so 1 times DX that's going to be DX is 2 cosine theta D theta so let me write that that is 2 cosine theta D I'll write the D theta out here and then what's the denominator square root of 4 minus x squared well that's 2 cosine theta again so that is two cosine theta well this seems to work out quite nicely if you to cosine theta over two cosine theta that's just going to be 1 this simplifies this simplifies to D theta which is if you just evaluate this this is just going to be equal to theta plus C well this is kind of nice but we're still not done we want this we want our indefinite integral in terms of X so now let's just solve for X here so if X is 2 cosine theta or sorry if X is 2 sine theta so X is equal to 2 sine X is equal to 2 sine theta then let's see divide both sides by 2 x over 2 is equal to sine theta and then if you want to solve for theta theta is the angle that if you take the sine of it you get x over 2 so we could say let's give ourselves another a little bit more real estate that theta is equal to the inverse sine the inverse sine of this thing x over 2 we could write it that way or we could write that theta is equal to the arc sine the arc sine of the arc sine of X over 2 so this is going to be equal to the art theta is arc sine of X over 2 so let's just do that arc sine of X over 2 plus C and we're done we've just evaluated that indefinite integral now some of you might have noticed something I kind of brushed past it really fast just to give you the the Big C though see the forest for the trees but there's some interesting details and it's I think worth digging into a little bit over here so the first one is you might see that look the restriction on X here is or the domain here on this expression is restricted so let's just keep track of that to make sure that we didn't do anything strange when we did that substitution so the domain here X has to be X has to be greater than negative 2 and less than 2 if the absolute value of X were equal to 2 then you would have a 0 and nominator if the absolute value of X is greater than 2 then you're going to have a negative in the denominator and that's not defined so this right over here is the domain so let's say make sure that our substitution didn't do anything weird with that so if X has to be between negative 2 and 2 and we're saying X is 2 sine theta that means 2 sine theta would have to be between negative 2 & 2 so negative 2 would have to be less than 2 sine theta 2 sine theta which would have to be less than 2 we could divide all the different parts of this compound inequality by 2 and you're going to get negative 1 is less than sine theta is less than sine theta which is less than 1 and the way we can do that is if theta is less than PI over 2 at PI over 2 sine of theta would be equal to 1 and if theta is greater than negative PI over 2 so if we restrict it in this way if we say theta is going to be in this range right over here then then where we are restricting our domain in a reasonable way and this works out well because this typically is the the range for the arc sine function so we could feel good about that now another question that you might have is okay well look you know we divided by cosine of theta here but that's okay as long as cosine of theta does not equal 0 because you don't want to zero in the denominator and the good thing about this restriction on theta is as long as theta is greater than negative PI over 2 and less than PI over 2 cosine of theta is going to be it's not going to be 0 and actually it's going to be positive if if negative PI over 2 or pi over 2 are allowed then you would get a 0 down here and we would have to think about restricting things in some other ways so it looks like everything is cool we we dug a little deeper we said okay we haven't done anything strange to the domain or to kind of unrestricted it in some strange way and so we can feel we can feel good about we can feel good about this about this this this answer that we've gotten