U-substitution
Another u-substitution example Finding the antiderivative using u-substitution.
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- 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.
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