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let's say we have the indefinite integral of the square root of 6x minus x squared minus 5 and obviously this is not some simple integral I don't have just you know this expression and then it's derivative lying around so u substitution won't work and so you can guess from just the title of this video that we're gonna have to do something fancier and we'll probably have to do some type of trig substitution but this immediately doesn't look kind of amenable to trig substitution I like to do trig substitution when I see kind of a 1 minus x squared under a radical sign or maybe an x squared minus 1 under a radical sign or maybe a x squared plus 1 these are the type of things that get my brain thinking in terms of trig substitution but that doesn't quite look like that just yet I have a radical sign I have some x squared but it doesn't look like in this form so let's see if we can get it to being in this form so let me delete these guys right there real fast so the the interest let's see if we can write maybe complete the square down here so let's see if this is equal to let me rewrite this and if this completing the square doesn't look familiar to you have a whole bunch of videos on that let me rewrite this as equal to minus 5 - I need more space up here minus 5 minus x squared now there's a plus 6 X but I have a minus out here so minus 6 X right a minus and a minus will become this plus 6 X and then I want to make this into a perfect square so what number when I add to itself will be minus 6 well it's minus 3 and minus 3 squared so you take half of this number you get minus 3 and then you square it and so you put a 9 there now I can't just arbitrarily add 9 so actually I didn't add a 9 here what did i do i subtracted a 9 because I threw a 9 there but it's really a minus 9 because of this minus sign out there so in order to make this neutral to my 9 that I just threw in there this is a minus 9 I have to add a 9 so let me add a 9 so plus 9 right there if this doesn't make complete sense what I just did and obviously have the DX right there multiply this out multiply this you get minus x squared plus 6x which are these two terms right there minus 9 and then this +9 and these two will cancel out and you'll just get exactly back to what we had before because I want you to realize I didn't change the equation and this is a minus 9 because of this so I added a 9 so I really added zero to it but what this does is it gets it into a form that I like obviously this right here just becomes a 4 and then this term right here becomes what that is X minus 3 squared X minus 3 squared so my indefinite integral now becomes the integral I'm just doing a little bit of algebra the integral of the square root of 4 minus 4 minus X minus 3 squared X minus 3 squared DX now this is starting to look at like a form that I like but I like to have a 1 here so let's factor a 4 out so this is equal to I'll switch colors that's equal to the integral of the radical and we'll have the 4 times 1 minus X minus 3 squared over 4 I just took a 4 out of both of these terms if I multiply this out I'll just get back to that right there DX and now this is trying to look like a form that I like let me simplify it even more so this is equal to the integral if I take the square 4 out it becomes 2 times the square root of 1 minus and I can rewrite this as X minus 3 let me write it this way 1 minus X minus 3 over 2 squared DX and where did I get that to from well if I just square both of this I get X minus 3 squared over 2 squared which is X minus 3 over 4 so I've just just I have done no calculus so far I just algebraically rewrote this indefinite integral as this indefinite integral they are equivalent but this all of a sudden looks like a form that I recognize I showed you in the last video that cosine squared of theta is just equal to 1 minus sine squared of theta you could actually do it the other way you can do sine squared is equal to 1 minus cosine squared no difference but they both will work out but this this looks an awful lot like this in fact it look exactly like this if I say that that is equal to sine squared of theta so let me make that substitution let me write let me write that X minus three X minus three over two squared is equal to is equal to sine squared of theta now if we take the square root of both sides of that equation I get X minus three over two is equal to sine of theta now where eventually you know where this is going to go we're eventually gonna have to substitute back for for theta so let's solve for theta in terms of X so theta in terms of X we could just say just take the arc sine of both sides of this you get theta is equal to write the arc sine of the sine is just theta theta is equal to the arc sine of X minus three over two fair enough now to actually do the substitution though we're going to have to order figure out what DX is we're going to solve for x in front terms of theta so let me do that so we get if we multiply both sides equation by 2 we get X minus 3x minus three is equal to two times the sine of theta or that X is equal to two sine of theta plus three if we take the derivative of both sides with respect to theta we get DX D theta is equal to two cosine of theta derivative this is just zero or we can multiply both sides by D theta and we get DX is equal to two cosine of theta D theta and we're ready to substitute back into our original indefinite integral so this thing will now be re-written as the integral of two times the radical and we get some space of one - I'm replacing this with sine squared of theta sine squared of theta and then all of that times DX well you just said that DX is equal to this right here so DX is equal to to cosine of theta d-theta now what does this simplify to this action right here this is just cosine squared of theta and we're going to take the square root of cosine squared of theta so this is the square root of cosine squared of theta this whole term right here right that becomes the square root of cosine squared of theta which is this the same thing as cosine of theta so our integral becomes so our integral is equal to two times the square root of cosine squared of theta so that's just two times cosine of theta times two times cosine of theta times two times cosine of theta that's that one right there this is this and all of this radical sign that's this right there it just this one minus sine squared was cosine squared take the radical you get cosine squared and then everything times D theta D theta now this is obviously this is equal to four times cosine squared of theta D theta which by itself is still not an easy integral to solve I don't know I you know I it's I can't do u substitution or anything like that there so what do we do well we resort to our good old our good old trig identities now I don't know if you have this one memorized although it tends to be in the inside cover of most calculus books or inside cover over most trig books but cosine squared of theta cosine squared of theta can be re-written as one half times one plus cosine of two theta and I've proven this in multiple videos so let's just make this substitution let me just replace this thing with that thing so this integral will become it becomes it equals four times cosine squared of theta but cosine squared of theta is this four times one half times one plus cosine of two theta D theta d-theta now this looks easier to to deal with so what is this 4 times 1/2 that's 2 so my integral my integral becomes the integral of 2 times 1 so it's 2 plus 2 times 2 cosine of 2 theta all of that D theta now this antiderivative is pretty straightforward what is this right here this is this is the derivative with respect to theta of cosine of sorry of sine of 2 theta right this whole thing what's the derivative of sine of 2 theta take the derivative of the inside that's 2 times the derivative of the outside cosine of 2 theta and this of course is the derivative of 2 theta so this is equal to the antiderivative of 2 with respect to theta is just 2 theta plus the antiderivative of this which is just sine of 2 theta sine of 2 theta sine of 2 theta and then we have a plus C and of course we can't forget that we defined theta we you know our original antiderivative was in terms of X so we can't just leave it in terms of theta we're gonna have to do a back substitution so let's just remember theta was equal to arc sine of X minus 3 over 2 let me write that on the side here theta is equal to arc sine of X minus 3 over 2 now if I immediately substitute this data straight into this I'm going to get a sine of 2 times arc sine of X minus 3 over 2 which would be correct and I would have a 2 times arc sine of X minus 3 over 2 and that would all be fine and we would be done but that's not satisfying it's not a nice clean answer so let's see if we can if we can simplify this so it's only in terms of sine of theta because you take the sine of the arc sine then you're definitely then it just simplifies to X minus 3 over 2 let me let me make that clear if I take so if I can write all of this in terms of sines of theta because the sine of theta let me write that the sine of theta is equal to the sign of the arc sine sine of the arc sine of X minus three over two which is just equal to X minus three over two so if I can write this in terms of sines of theta then I can just make this substitution sine of theta equals that and everything simplifies a good bit so let's see if we can do that well you may or may not know the the other identity and I've proven this as well that sine of two theta sine of two theta that's the same thing as sine of theta times sine of theta or the sine of theta plus theta that's equal to sine of theta plus theta same thing which is equal to sine of theta cosine of theta plus the other Thetas get swapped around sine of theta plus cosine of theta which is equal to this is just the same thing written twice to sine of theta cosine of theta some people have this memorized ahead of time and if you're going to take an exam on trig substitution doesn't hurt to have this memorized ahead of time but let's rewrite this like this so our indefinite integral in terms of theta or our antiderivative became 2 theta plus instead of sine of two theta we could write 2 sine of theta cosine of theta and then of course we have a plus C now I want to write everything in terms of sines of theta but I have a cosine of theta there so what can we do well we know we know that cosine squared of theta is equal to one minus sine squared of theta or that cosine of theta is equal to the square root of one minus sine squared of theta which seems like we're adding complexity to it but the neat thing is it's in terms of sine of theta so let's do that let's make the substitution so our antiderivative this is the same thing as two theta plus 2 sine of theta times cosine of theta which is equal to this times the square root of 1 minus sine squared of theta and then all that plus C and now we're at the home stretch this problem was probably harder than you thought going to be we know that sine theta is equal to X minus 3 over 2 so let's make that reverse substitution so we have 2 times theta this first term is just 2 times theta right there so that's 2 times we know what theta is equals we can't escape the arcsine if you just have a theta we have to say that theta is equal to arc sine arc sine of X minus 3 over 2 and then we have plus let me switch colors plus 2 sine of theta well that's plus 2 times sine of theta is X minus 3 over 2 so 2 times X minus 3 over 2 and then all of that times the square root the square root of 1 minus sine of theta squared what's sine of theta it's X minus 3 over 2 squared and of course we have a plus C and let's see if we can simplify this even more now that we're already at the homestretch so this is equal to 2 arc sine of X minus 3 over 2 plus these two terms that this 2 and this 2 cancel out plus X minus 3 times the square root of what happens if we multiply everything here by let's see if we let's see if we take a so this is equal to 1 minus X minus 3 over 4 X minus 3 squared over 4 this simplification I should write that in quotes is taking me longer than I thought but let's see if we can simplify this even more if we multiply let me just focus on this term right here if we multiply the outside if or let's say let's let's multiply and divide this by 2 so I'll write this as so let's just multiply this times 2 over 2 2 over 2 and you might say Sal why are you doing that because I can rewrite this let me write my whole thing here so I have to arc sine of X minus 3 over 2 and then I have I can take this denominator 2 right here so I say plus X minus 3 over 2 that's that too is that 2 right there and then I could write this 2 right here as the square root of 4 times the square root of 4 times the square root of all of this 1 minus X minus 3 squared over 4 and I think you see where I'm going I'm kind of reversing everything that I did at the beginning of this problem and maybe I'm getting a little fixated on making this as simple as possible but I'm so close so let me finish so I get 2 times the arc sine of X minus 3 over 2 which I am tired of writing plus X minus 3 over 2 and then if we bring this 4 in write the square root of 4 times the square root of that is equal to the square root of 4 times these things so it's 4 minus X minus 3 squared all of that plus C and we're at the home stretch this is equal to 2 times the arc sine of X minus 3 over 2 plus X minus 3 over 2 times the radical of 4 minus let's expand this x squared minus 6x plus 9 and then this expression right here simplifies to this simplifies to minus minus it's 6 X 6x minus x squared and then you have a 4 minus 9 minus 5 which was our original antiderivative which is our original antiderivative so finally we're at the very end we get the antiderivative is to arc sine of X minus 3 over 2 plus X minus 3 over 2 times let me make this times the radical of 6x minus x squared minus 5 that right there is the antiderivative of what this thing that we had at the very top of my little chalkboard which is right there so that is equal to the antiderivative of the square root of 6x minus x squared minus 5 DX and I can imagine that you're probably as tired as I am my hand actually hurts but hopefully you find that to be vaguely satisfying sometimes I get complaints that I only do easy problems well this was quite a hairy and not so easy problem