Evan from Norway has asked me to do another u substitution
problem, and I like these because it gets my momentum
going for doing other things that maybe take a little
bit more preparation.
And the problem he sent me-- and I hope I pronounced his
name right-- was the indefinite integral of sin of x over
the cosine of x squared dx.
This it also be written as-- he wrote it in email, so I don't
know how he exactly saw it, but it can also be written as sin
of x over cosine squared of x.
Sometimes it's written like that.
Either way, I like looking at this a bit better.
It's a little bit less ambiguous.
But in general you know to do u substitution, or integration by
substitution when you see something and you see its
derivative sitting there.
Right.
You're like wow, this cosine of x was just an x, or if it was
just a u, this would be a really easy integral to do.
We know how to do this integral.
Let me do it on the side.
This integral would be easy.
1 over x squared dx.
We know how to do that.
That would just be the antiderivative of x squared.
This is the same thing as the antiderivative of x to the
minus 2 dx, and we know how to take the antiderivative
of something like that.
You increase the exponent by 1 and then you multiply by
what your new exponent is.
So it would be minus x-- I'm sorry.
You increase your exponent by 1, and then you divide by
whatever your new exponent is.
So you would, [? let me do the, ?]
increasing the exponent x to the minus 2, you'd increase
the exponent, you'd get x to the minus 1.
And then when you divide this by minus 1 you get this minus
out front, and then of course you'd have the plus c.
If you don't believe it take the derivative.
Negative 1 times minus 1.
That's a positive.
And then you'd decrease the exponent by 1, you
get x to the minus 2.
So if we could get it in a form that looks like
this, we'd be all set.
And you kind of do see a pattern.
Where this x is you have a cosine there, and then we have
cosines derivative there.
So that's the big clue that we should be using u substitution.
So let's do that.
And what we're going to do is we're going to substitute
u for cosine of x.
So if we say u is equal to cosine of x, and let's take
the derivative of u with respect to x.
So du/dx is equal to what?
What's the derivative of cosine of x? it's not
quite sin of x, right?
It's minus sin of x.
And then we can multiply both sides by dx, and you get du is
equal to minus sin of x dx.
I just multiplied both sides by dx.
And then up here we have sin of x dx.
We don't have minus sin of x dx.
There we have sin of x dx.
We could have rewritten this top integral, we could have
rewritten it like this.
Sin of x dx.
All of that over cosine of x squared.
So if we want to substitute for this, here we have a minus.
Let's multiply both sides of this by a negative 1, and you
get minus du is equal to sin of x dx.
And let's see.
So let's rewrite this original problem.
Now I know I'm running out of space.
Let me rewrite it.
So we know that u is equal to cosine of x, so let's do that.
So now this integral becomes-- and the denominator, instead of
cosine of x squared, u is cosine of x.
That's u, right?
We made that definition.
So that's over u squared.
Cosine of x becomes u.
And then sin of x dx right up there, what is that equal to?
Well we just solved for it here.
That's equal to minus du.
Sin of x dx is equal to minus du.
So that we can replace with this, minus du.
And then of course this has the exact same form as
this thing right here.
You could rewrite this, this is equal to let's say
minus 1 over u squared du.
I'm just writing it a bunch of different ways.
Whatever is easier for you to conceptualize.
The same thing as minus u to the minus 2 du.
And then here we do the same thing we did up here, although
now we have a minus out front, that actually makes it
a little bit cleaner.
To take the antiderivative, we raise u-- it was to the minus 2
power, let's raise it to 1 power higher than that-- so
minus 2 plus 1 is minus 1.
So it's u to the minus 1 power, and then you want to divide by
minus 1, and I'll do it explicitly here.
Minus one.
And then you had this negative that was sitting out there
before, so that negative is still going to be there.
And of course you're going to have a plus c.
You can view this as a negative 1 or, this negative divided
by a negative, they're going to cancel out.
And so you're just left with u to the minus 1 plus c, or 1
over u plus c is the antiderivative-- oh sorry,
we're not done yet.
That's just the antiderivative of this.
And now we have our substitution to deal with.
What was our substitution that we started with?
u is equal to cosine of x.
So if u is equal to cosine of x, this thing is equal to 1
over cosine of x plus c is equal to the antiderivative of
our original problem, which was sin of x over cosine
of x squared dx.
There you go.
See you in the next video.