in this video i want to focus on some different integration techniques that you'll learn in calculus 2. so let's say if we wish to integrate this function x sine x dx how can we do so for a situation like this notice that you have two different types of functions being multiplied by each other a linear function x and a trigonometric function sine x when you see two different components being multiplied like this you could use something called integration by parts and the formula goes like this the integration of u dv is equal to u times v minus the integral of v d u so we need to know what u and what dv is what we're going to do is we're going to make u equal to x so that means that d u is going to be the derivative of x which is 1 times dx or simply dx now dv is going to be sine x with the dx as well so if we integrate both sides the integral of dv will give us v and the integral of sine is negative cosine x now that we have this we can use the right side of this equation so let's go ahead and plug in everything so the integral of x sine x dx is going to be u times v so we know u is x v is negative cosine x so that's going to be negative x cosine x and then minus the integral of v d u so v is still negative cosine x and d u is equal to dx so at this point we no longer need this information so let's go ahead and get rid of that and let's simplify the expression that we now have so we have two negative signs which we can make that a positive sign and the anti-derivative of cosine x we know it's sine x and of course anytime you have an indefinite integral you need to include the constant of integration plus c so this is the answer it's negative x cosine x plus sine x plus c and that's it now let's try another example using integration by parts find the anti-derivative of the natural log of x times dx feel free to pause the video if you want to give this problem a shot go ahead and try it now we need to determine what we're going to make u and dv equal to so keep in mind the formula is the integral of u dv is equal to u times v minus the integral of vdu so there's two parts here we have dx and ln x so what should we do in this case now to find the integral of l and x is what we're trying to do so it makes no sense to set dv equal to ln x instead we need to make u equal to ln x and dv has to be equal to dx now we need to find d u the derivative of l and x is going to be 1 over x times dx i know this kind of looked like a v but let me just redraw that as a u now v is going to be the integral of dx so if dv is dx v has to be equal to x now let's apply the formula that we have here so the integral of ln x dx is going to be u times v and so we can see that u is l and x v is x so u times v is going to be x ln x and then minus the integral of v which is x times d u and d u is 1 over x dx so let's get rid of that and now let's simplify the expression that we currently have so x times one over x is one the x variables will cancel so we have the integral of 1 times dx or simply dx the integral of dx is x and don't forget to add plus c so this is the answer right here the integral of ln x is x ln x minus x plus c now the next topic we're going to go over is trigonometric integrals so how can we find the integral of cosine cubed x dx go ahead and take a minute on that one what do you think we need to do so basically we need to do u substitution but we need to adjust the problem that we have for instance cosine q we can write that as cosine squared times cosine now if you recall one of the pythagorean identities for trigonometry is this one sine squared plus cosine squared is equal to one and so if you solve for cosine squared by moving sine squared to the other side you'll find that cosine squared can be replaced with 1 minus sine squared and so that's what we're going to do here we're going to replace cosine squared with that so we're going to have 1 minus sine squared times cosine and now in this form we could use u substitution so let's make the u variable equal to sine x if we do so d u is going to be the derivative of sine which is cosine x times dx so now let's replace sine with the u variable and let's replace cosine x dx with du and now we can integrate this this is supposed to be uh u squared so the antiderivative of one d u is u and the antiderivative of u squared is u to the third over three using the power rule now the last thing we need to do is replace the u variable so the final answer let's see if i can fit it here somewhere maybe i'll just have to get rid of this the final answer is going to be sine x minus 1 3 sine cube x plus c and that's it that's all you need to do in this problem now here's another problem let's find the integral of cosine to the fifth power times sine to the fourth power so it's somewhat similar to the last problem but it's a little bit longer feel free to try if you want to now notice that the trig function cosine on the left is raised to an odd power and the sine function is raised to an even power we want to focus on cosine because it's raised to an odd power we can split it into two components one that's odd and when that's even so cosine to the fifth power what we want to do is break it up into cosine x times cosine to the fourth power the even portion we can convert it into sine and so that's going to help us when we use u substitution now the first thing we're going to do is replace cosine to the fourth with cosine squared raised to the second power because 2 times 2 is 4. and now just like before we can replace cosine squared with one minus sine squared now we're ready to use u substitution so just like before we're going to make u equal to sine x and d u is going to be cosine x dx so everywhere we see a sine function let's replace it with the u variable so this is going to be 1 minus u squared raised to the second power times u to the fourth and then cosine x times dx all of that we can replace with du so now at this point what we need to do is foil this expression so let's multiply 1 minus u squared by itself before we distribute the u to the 4. and so 1 times 1 that's going to be 1 and then we have 1 times negative u squared and then negative u squared times 1 and finally negative u squared times negative u squared which is positive u to the fourth now our next step is to combine like terms negative u squared minus u squared that's going to be negative two u squared now let's delete this and let's distribute the u to the fourth to everything inside the brackets so u to the fourth times one will be the same thing and then u to the fourth times negative two u squared two plus four is six and then four plus four is eight now let's use the power rule in the next step so it's gonna be u to the fifth over five minus two u to the seven over seven plus u to the nine over nine now the last thing we need to do is replace u with sine x so the final answer is going to be one over five sine to the fifth x minus two over seven and then sine to the seventh x plus one over nine sine to the ninth x plus c and so that's it for this problem by the way for those of you who want more examples on integration by parts or trig integrals feel free to check out my new playlist my new calculus video playlist and i'm going to post the link in the description section so you can take a look at that when you get a chance but in this video i'm just briefly going over the main integration techniques that you need to be familiar with in calc 2 for those of you who are going to take it now here's another problem for you what is the integral of sine squared x dx try that one now there's no point replacing sine squared with 1 minus cosine squared it won't help us in this particular problem we can't really use u substitution so we have to do something different and this is where you need to know your trig identities particularly the half angle identity you need to know that sine squared is one half times one minus cosine two x and so we could replace sine squared with that now we can move the one half to the front and so we can rewrite the integral like this now at this point we can integrate the function so the anti-derivative of one dx is x and the anti-derivative of cosine is sine but the anti-derivative of cosine two x is sine two x but divided by two and i'll explain why but for now let's finish this problem so i'm going to distribute the one half so it's going to be one half x minus and then a half times another half that's a fourth or two times two is four so this is going to be one fourth sine two x plus c and so this is the answer for this problem now for those of you who want to see why the integral of cosine 2x is one half sine two x you need to use u substitution if you make u equal to two x d u is gonna be two times d x solving for d x it's d u divided by two and so what we're going to do is we're going to replace 2x with the u variable and so this becomes cosine of u and then let's replace the dx with du over 2. now let's move the two to the front so it's one half integral of cosine u now the anti-derivative of cosine is positive sine and then the last thing we need to do is replace u with two x and so that gives us this answer which is one half sine two x plus c so the antiderivative of cosine 2x as you could see is sine 2x divided by 2. now let's try a problem with tangent and secant so let's try this one let's find the anti-derivative of tangent to the sixth power times secant raised to the fourth power now some things you want to keep in mind the derivative of tangent is secant squared and also make sure you know this particular trig identity one plus tangent squared is secant squared so using that information what is the antiderivative of tangent to the sixth power times secant to the fourth power now we need to make u equal to tangent so that d u is going to be secant squared dx if we made u equal to let's say secant x d u will be secant tangent and that will make the situation a lot more complicated than it needs to be so we're not going to do it that way now we know that the secant squared will disappear but we have secant to the fourth so therefore we need to get rid of a secant squared in order for this to work so let's split up secant to the fourth power into secant squared times secant squared so this portion secant squared dx we're going to replace it with d u in time but somehow we need to get rid of this secant squared now recall that we said that secant squared was 1 plus tan squared so now it's a good time to use that particular identity so let's replace secant squared with one plus tan squared so now at this point we can substitute everywhere we see a tangent we're going to replace it with the u variable and wherever we see uh secant squared dx which is only here once we're going to replace that with the u so now we need to distribute u to the six to one and u squared so this becomes the integral of u to the six plus u to the eighth times d u now using the power rule the antiderivative of u to the sixth is u to the seven over seven and for u to the eight it's gonna be u to the nine over nine plus c so now let's replace the u variable with the tangent and so the final answer is gonna be one over seven tangent to the seventh power of x plus one over nine tan raised to the nine power plus c and that's it so this is the anti-derivative of tangent to the sixth power times secant to the fourth power now there's something called trigonometric substitution another integration technique that you're going to learn in calc 2 and whenever you see an integration problem in this form a squared minus x squared inside a square root symbol you need to set x equal to a sine theta and the reason for this is because 1 minus sine squared is cosine squared if you see this form a squared plus x squared within a root symbol your substitution will be this x is equal to a tangent theta and finally if you see the square root of x squared minus a squared you need to set x equal to a secant theta because secant squared minus 1 is tan squared so based on that let's try a problem let's find the antiderivative of the square root of 4 minus x squared divided by x squared dx so we could see that we have a problem in this form and so 4 is equal to a squared therefore a is equal to 2. now x has to be equal to a sine theta so in this problem x is 2 sine theta which means that dx is going to be 2 times the derivative of sine which is cosine theta and then d theta now x squared is going to be 2 sine theta squared which is 4 sine squared theta now what we're going to do is we're going to replace x squared with 4 sine squared theta and at the same time we're going to replace dx with 2 cosine theta d theta so now inside the square root let's take out a 4. so if we factor out the gcf we're gonna have one minus sine squared on the inside now i'm going to run out of space so i'm going to erase everything on top so just keep this in mind x is equal to 2 sine theta now the square root of four is two so we can take out the two from the radical and one minus sine squared is equal to cosine squared so this is what we now have in the next step we can take the square root of cosine squared and so that's i didn't want that to happen that's going to be cosine and so we're going to have 2 times cosine theta times another 2 cosine theta d theta over 4 sine squared theta now 2 times 2 on top is 4 which will cancel with the 4 on the bottom and so we're left with cosine times cosine which is cosine squared and cosine squared divided by sine squared is cotangent squared so what do you think we need to do at this point what's our next our next step here excuse me now recall that 1 plus tangent squared is secant squared so therefore one plus cotangent squared is cosecant squared moving the one to the other side we can replace cotangent squared with cosecant squared minus one and so this is what we now have the integral of cosecant squared theta minus one d theta now what is the antiderivative of cosecant squared you need to know that the derivative of cotangent is negative cosecant squared and so the derivative of negative cotangent is positive cosecant squared so the antiderivative of positive cosecant squared theta is negative cotangent theta and the integral of negative one d theta is simply negative theta so right now we're getting close but we don't have the answer quite yet even though we've integrated or we found the anti-derivative of this expression we still need to replace theta with some x variable as you can see this is a very long problem so let's go back to this expression where x is 2 sine theta let's divide by two so sine theta is x over two and what we're going to do is we're going to make a right triangle and let's do that a little bit better and so here's theta now based on sohcahtoa we know that sine is equal to the opposite side divided by the hypotenuse so x goes on the opposite side two is the hypotenuse now using the pythagorean theorem we need to find a missing side so a squared plus b squared is equal to c squared and let's say a is the missing side that means b is equal to x so b squared is x squared c squared is two squared so moving x squared to the other side we'll get that a squared is 4 minus x squared and taking the square root of both sides we can see that the missing side is the square root of 4 minus x squared so now we can figure out what tangent is based on sohcahtoa tangent is opposite over adjacent so it's x over the square root of 4 minus x squared cotangent is the reciprocal of tangent so cotangent is going to be square root 4 minus x squared over x and we do have a negative sign now what about theta what is theta equal to well we saw that sine theta is x over two so theta has to be arc sine x over two and so we can replace this theta with that so we have minus arc sine x over two and then plus the constant of integration so this is the final answer so that's an example of using trigonometric substitution to find the antiderivative of something now because it's a long problem i'm only going to do one example in this video if you want more examples check out my new calculus video playlist in the description section of this video or you can subscribe to this channel and check out all the other video content that i have thanks again for watching