okay so let's do two more examples let's do this one the integral of e to 2x * sin of 3x okay this example I admit is annoying so we're going to choose a u and a DV unfortunately neither of these U's simplifies tremendously when I take a derivative of U so I'm going to just pick either one let's say I pick U as e 2X and that means my DV is s of 3x well at least I have a DV that I know I can find V so that's good that's a partial success so du is exponentials are their own derivatives but I have chain rules so it's 2 e to 2x DX and V the integral of s of anything is minus cosine and I have to get rid of that problem with the three so- 13 cosine of 3x okay so my first integration by part step is the integral of U DV is U * V minus the integral of vdu so it's U * V which is these guys I'll put the constant in front - 13 e 2x * cosine of 3x okay minus the integral of V du which is this so there's a 2 and a minus 1/3 and this is minus so minus and minus becomes plus and I'm going to put that two and that 1/3 together out here and then I have e to 2x 3x DX the bad news is this doesn't look any simpler than what I started with in fact it's really similar to what I started with except I have cosine instead of s so here's a trick that you might not think of if you hadn't done it before which is we are going to just use integration by parts again okay I'm going to kind of ignore this 2/3 for a minute so I'm going to rewrite minus uh let me rewrite this part in blue I'm going to rewrite this first part minus 1/3 so this equals minus 1/3 I'm going to just rewrite this part plus 2/3 and here I'm going to do kind of like I just did a minute ago I'm going to use integration by parts on this inside guy Okay so so my U here's a trick for you um it's useful to pick the same type of U that you picked in the beginning so if I picked an exponential here I could have just as easily picked U to be S of 3x but since I picked U to be e to 2x I'm going to pick the same thing in this integral here so I'm going to pick U to be e to 2x again and my DV my DV is going to be the DV for this integral so it's going to be this cosine it's going to be cosine of 3x DX so my du is 2 e 2x DX and my V is a function whose derivative is cosine is s and I have to deal with that three inside so when I'm taking anti-derivatives I have to divide by three okay so inside here I'm going to just put my integration by parts formula so I have U * V I'll put the 1/3 out front 1 E 2x sin of 3x minus the integral of V du the constants 23 I'll put those out front 23 V du I'll put the e to the 2x first I'll do U DV s e 2x s of 3x DX so bummer cu this is the same as this so I just did a giant Circle okay so hm but actually it's good that this is the same as this CU everything else is calculated that means I can solve for this guy okay so here's what I'm going to do I'm going to rewrite everything that I have just to see it in one line e 2x sin of 3x DX equals this plus this plus minus this so it's equal to - 13 e 2x cosine 3x plus 2/3 and 1/3 is 2 9th e 2x sin of 3x okay and then there's a minus there's a 2/3 and a 2/3 2/3 * 2/3 is 4 and 9th and I have a minus - 4 9th and here I have an integral okay do is I'm just going to solve for this guy I'm just going to take this guy and add it to the other side I have one of these yellow integrals and when I do add this I'm going to have plus 4 9ths of this same integral equals this thing that I can calculate I have to be a little careful because this involves a plus c and this involves like this involves like a plus C1 and this involves a plus C2 so when I put these integrals together I'm going to add a plus c but so what happens I have this integral when I move this to the other side plus 49 time this integral so I have 1 + 4 9 * this integral equals well just equals what I have here with the plus C I have to add it in here because I kind of had it here before it was sort of hidden I didn't write it explicitly it's hidden in that notation of the indefinite integral so this is what I have let's C so this integral that I'm looking for is just what I have on this right hand side divided 1 + 4 9ths okay and what is 1 + 4 9ths 1 + 4 9ths is uh 9 + 4 is 13 9th okay so you have to check me I do make algebra mistakes so you check me on these if I made an algebra mistake you can fix it for yourself don't get dist stracted you can make fix it we can check me but this is 13 9th so in the end this integral equals 9 over 13 times what I have on this right hand side okay plus okay technically it's 913 time C but Conant anyway so I'll just write plus C1 maybe just to have an arbitrary constant okay and if I were really really if I cared a lot I could simplify this a little bit this nine and this three would simplify a bit but nobody really cares okay so when I have kind of these circular functions that when I differentiate um I basically uh get back to something similar when I take an anti- derivative twice here for example here I took an anti-derivative once I got back to cosine and I took an anti-derivative again I got back to S this type of technique works when both of the functions are are are of that type okay so that's a good technique to have at your disposal okay last example you have an integral of arc S of 3T you have an example like this but with arc tangent let's say I have t instead of X again this is a scenario where I don't have anything written in terms of X but also where I only have one choice for the function U so DV is just DT all right so we have to dig a little to remember what the derivative of this thing is so derivative of R sin of X is 1 < TK of 1 - x^ 2 here I have t and my X is 3 T so when I square that the 3^ S becomes a 9 and the t^2 and you have a chain Ru because there's a three here so actually this one I'm going to put it up here becomes a 3 DT okay so that's the derivative of R of 3T using what the derivative of AR sign is and the chain Rule and if DV is DT and this is just like a one that means that V is just T because T is one so that means this is a good choice all right so remember integral u d v is U * V minus integral vdu so this is U * V for some reason we put the T first but it doesn't really matter it's just kind of tradition we mathematicians are kind of traditional and the integral of V duu so I'm going to take this T and multiply it by this so it's like multiplying by T over one so I'm just going to put the 3 t on the numerator okay okay now depressed looking at this because you might be like hm can I integrate that well yes you can but you need to do a u substitution here so this is mixing not a second integration by parts but doing you substitution this is what you were already able to do last year in calc one substitution okay so this is a I'm going to write U but it's a different U than integration by parts U okay this is U substitution you so what's a good choice for you here you need something who's derivative shows up and such that U simplifies this integral so here's a good choice for you U is 1 - 9 t^2 so du is - 9 2 - 18 T DT and different people do this uh differently um I kind of like to see oh what do I have in my integral I have 3T DT if I want I could even put this minus in there and how can I get a minus 3T DT here well I have a minus I have a three in this 18 it's there's a six too of too much that's too much so6 du is - 3 tdt okay so this is -3 * 6 tdt - 3 TD what I have is the only thing missing okay if you have a different way of doing this then feel free to use your Technique here okay this is just the way I like to see it because now I'm going to replace what I've highlighted in yellow with exactly 1 16 du so instead of a minus I'm going to have a plus and I'm going to have 1 16 so this guy equals this is the same t r sin 3T - 3T DT became + 16 this is 1/ square root of U du and now I have to integrate U 1/ Square < TK of U is the same as square root of U is U to the 1/2 but since it's on a denominator it's U to the minus one2 so what is is the integral of U to the -2 I add one right so it's U to the 12 over 12 or 2 < TK U so the integral of 1 < TK of u d is 2 < TK U + C and instead of U I'm going to write what my U is I'm just going to do that right away C and this is basically my answer now of course I could take this two and the six and write 1/3 instead okay so I probably would write one last step replacing this by three okay