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Let's see if we can take the integral of cosine of 5x over e to the sine of 5x dx.
And there's a crow squawking outside of my window so I'll try to stay focused.
So let's think about whether u-substitution might be appropriate. Your first temptation might to said,
"Hey, maybe we let u equal sine of 5x, and if u is equal to sine of 5x,
we have something that is pretty close to du up here." Let's verify that.
So du could be equal to -- so du/dx (derivative of u with respect to x),
well we just use the chain rule. Derivative of 5x is 5,
times the derivative of sine of 5x with respect to 5x, that's just going to be cosine of 5x.
If we want to write this in differential form, which is useful when we do our u-substitution,
we could say that du is equal to 5 cosine 5x.
Now when you look over here, we don't have quite du there. We have just cosine of 5x dx--
sorry, I need cosine of 5x dx, just like that. So when you look over here,
you have a cosine of 5x dx, but we don't have a 5 cosine of 5x dx,
but we know how to solve that. We can multiply by 5 and divide by 5.
1/5 times 5 is just going to be 1. So we haven't changed the value of the expression.
But when we do it this way, we see pretty clearly, we have our u and we have our du.
Our du is 5 -- let me circle that and let me do that in that blue color --
is 5 cosine of 5x dx. So we can rewrite this entire expression as --
I'll do that 1/5 in purple -- this is going to be equal to 1/5 --
I hope you don't hear that crow outside; he's getting quite obnoxious --
1/5 times the integral of, well all this stuff in blue is my du,
and then that is over e to the u. So how do we take the anti-derivative of this?
Well, you might be tempted to -- well, what would you do here?
Well, we're still not quite ready to simply take the anti-derivative here.
If I were to rewrite this, I could rewrite this as (this is equal to)
1/5 times the integral of e to the negative u du.
And so, what might jump out of you is maybe we do another substitution,
and we already use the letter u, so maybe we might use w. We'll do some "w-substitution."
And you might be able to do this in your head, but we'll do w-substitution just to make it a little bit clearer.
So let's -- this would've been really useful if this was just e to the u,
because we know the anti-derivative of e to the u. It's just e to the u.
So let's just try to get it in terms of the form of e to the negative something.
So let's set -- and I'm running out of colors here -- w equal to negative u.
And in that case, then dw (derivative of w with respect to u) is negative 1,
or if we were to write that statement in differential form,
dw is equal to du times negative 1 is negative du.
So this right over here would be our w, and do we have a dw here?
Well we just have du; we don't have a negative du there.
But we can create a negative du by multiplying this inside by negative 1,
but then also multiplying the outside by negative 1.
Negative 1 times negative 1 is positive 1; we haven't changed the value.
We have to do both of these in order for it to make sense.
Or I could do it like this. So negative 1 over here, and a negative 1 right over there.
And if we do it in that form, then this negative 1 times du --
that's the same thing as negative du -- this is this right over here.
And so we can rewrite our integral -- it's going to be equal to --
now it's going to be negative 1/5 -- trying to use the colors as best as I can --
times the indefinite integral of e to the -- well instead of negative u, we could right w.
E to the w. And instead of du times negative 1 or negative du, we can write "dw."
Now this simplifies things a good bit. We know what the anti-derivative of this in terms of w.
This is going to be equal to negative 1/5 e to the w, and then we might have some constant there,
so I just do a plus C. And now we just have to all of our un-substituting.
So we know that w is equal to negative u, so we could write that --
so this is equal to negative 1/5 -- I want to stay true to my colors -- e to the negative u,
that's what w is equal to, plus C. But we're still not done un-substituting.
We know that u is equal to sine of 5x. So we can write this as being equal to
negative 1/5 times e to the negative u, which is negative u is sine of 5x,
and then finally, we have our plus C. Now, there was a simpler way that we could've done this
by just doing one substitution. But then you kind of have to look ahead a little bit
and realize that it was not trivial to take -- not to bad to take your anti-derivative of e to the negative u.
The inside that you might of have although you shouldn't really hold yourself
when you feel too bad when you didn't see that inside.
We could've rewritten that original integral -- let me rewrite it --
it's cosine of 5x over e to the sine of 5x dx. We could've written this entire integral as being equal to
cosine of 5x times e to the negative sine of 5x dx. And in this situation, we could've said
u to be equal to negative of 5x, and say well, if u is equal to --
or negative sine of 5x, then du is going to be equal to negative 5 cosine of 5x,
and we don't have a negative 5 -- oh, dx, we don't have a negative 5 here,
but we can construct one by putting negative 5 there, then multiplying by negative 1/5,
and then that would've immediately simplified this integral right over here to be equal to
negative 1/5 times the integral of -- well, we have our du -- let me do this in a different color --
that's the negative 5 -- let me do it this way -- negative 5 cosine of 5x dx.
So that is our du -- I'm just changing the order of multiplication -- times e to the u.
This whole thing now is u this second time around. So if we did it this way, with just one substitution,
we could've immediately gotten to the result that we wanted. You take the anti-derivative of this --
I'll do it in one color now, just 'cause I think you get the idea -- this is equal to
negative 1/5 e to the u plus C. u is equal to negative sine of 5x,
so this is equal to negative 1/5 e to the negative sine of 5x plus C. And we're done.
So this one is faster; it's simpler, and over time, you might even start being able to do this in your head.
This top one, you still didn't mess up by just setting u equal to sine of 5x;
we just have to do an extra substitution in order to work it through all the way.
And I was able to do this video despite the crowing crow outside -- or squawking crow.