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Current time:0:00Total duration:5:47

- [Voiceover] Let's see if we
can evaluate the indefinite integral x over two times sine of two x squared plus two, dx. And try to pause the video and see if you can work
through it on your own. So, let's see what is going on here. So, I have this x over
two, and then I have sine of two x squared plus two. Now, if I were just taking
the indefinite integral of sine of x, that is pretty straightforward. The indefinite integral of sine of x. Well, we know that the
derivative of cosine of x is equal to negative sine of x. So if I were to take the
derivative of negative cosine of x, that's going to be positive sine of x. So this is just going to
be negative cosine of x. And I could have made that even clearer. I could have put a negative
here and then a negative here. And you see, well look,
the anti-derivative of negative sine of x is just
cosine of x, and then I have this negative out here,
negative cosine of x. But that's not what I have here. I don't have sine of x. I have sine of two x squared plus two. But then I have this other
thing with an x here, and so what your brain
might be doing, or it's good once you get enough
practice when your brain will start doing this, say
okay, this is interesting. This kind of looks like
the derivative of this. And this is, if we were to call this... I'm tired of that orange. If we were to call a... Why am I doing this,
color changing this... If we were to call this f of x. If two x squared plus two is f of x, Two x squared plus two is f of x. What is f prime of x? Well, then f prime of x, f prime of x is going to be four x. And this thing right over
here isn't exactly four x, but we can make it, we can
do a little rearranging, multiplying and dividing by a constant, so this becomes four x. What if, what if we were to... What if we were to multiply
and divide by four, so we multiply by four there
and then we divide by four, and then we take it out
of the integral sign. And even better let's take this
two out so let's just take. So, let's take the one half out of here, so this is going to be one half. And so I could have rewritten
the original integral as one half times one
fourth, so it's one eighth times the integral, times the integral of four x times sine of two x squared plus two, dx. Well, now this is interesting, because if this is f of x, if... If this business right
over here if f of x, so we're essentially
taking sine of f of x, then this business right over here is f prime of x, which is a
good signal to us that, hey, the reverse chain rule
is applicable over here. We can rewrite this, we
can also rewrite this as, this is going to be equal to one. We can rewrite this as, the
same things as one eighth... They're the same colors. I'm using a new art program,
and sometimes the color changing isn't as obvious as it should be. So one eighth times the
integral of f prime of x, f prime of x times sine, sine of f of x, sine of f of x, dx, throw that f of x in there. So, sine of f of x. And so when you view it
this way, you say, hey, by the reverse chain rule, I have... I have a function, and I have
its derivative here, so I can really just take the antiderivative
with respect to this. This is essentially what
we're doing in u-substitution. You could do u-substitution
here, you could set u equalling this, and then du
is going to be four x dx. But now we're getting a little
practice, starting to do a little bit more in our heads. So, what would this interval
integrate out to be? Well, this would be one eighth times... Well, if you take the
antiderivative of sine of f of x with respect to f of x,
well, we already saw that that's negative cosine of
x, so this is going to be times negative cosine, negative cosine of f of x. Negative cosine of f of x, negative cosine of f of x. Woops, I was going for the blue there. And then of course you have your plus c. So what is this going to be? Well, instead of just saying f pri.. Instead of saying in terms
of f of x, we just say it in terms of two x squared. This is going to be... Or two x squared plus two
is going to be one eighth. Times, I have a function
and I have its derivative, so I can really just
integrate with respect to that function, so
it's going to be times negative negative cosine of two x squared plus two, two x squared plus two, and then, of course, I have my... I keep switching to that color. I have my plus c, and of
course, I could just take the negative out, it would be
negative one eighth cosine of this business and then plus c. And we're done. We have just employed
the reverse chain rule. We could have used
substitution, but hopefully we're getting a little
bit of practice here. Hey, I'm seeing something
here, and I'm seeing it's derivative, so let me
just integrate with respect to this thing, which is
really what you would set u to be equal to here,
integrating with respect to the u, and you have your du here. This times this is du, so you're, like, integrating sine of u, du. I encourage you to try to
use u-substitution here, and you'll see it's the exact
same thing that we just did.