Definite integral with substitution Solving a definite integral with substitution (or the reverse chain rule)
Definite integral with substitution
- So I've been sent this definite integral problem and it seemed
- as good as any, and I think the key with this is just to
- see a lot of examples.
- So let's do it.
- This definite integral is from pi over 2 to pi of minus cosine
- squared of x times sin of x dx.
- So before we just chug through the math and do the
- antiderivatives and use the fundamental theorem of calculus
- to evaluate the definite integral, let's think about
- what we're even doing.
- So I've graphed this function right here, minus cosine
- squared of x times sin of x.
- And what we care about, we're defining the definite
- integral between pi over 2, which is roughly here.
- Let me see if I can make this a little bit bigger.
- So between pi over 2 which is right there, and between pi.
- So the definite integral of this function between here and
- here is essentially the area of the curve between the
- curve and the x-axis.
- And since the a curve is below the x-axis here, this area is
- going to be a negative number.
- So that gives us immediately an intuition.
- We should be getting a negative number when we evaluate this
- and just to prove this I actually typed it in up here.
- So let's now evaluate this definite integral.
- Now I'll rearrange some of the terms here just to make it a
- little bit easier to read.
- But the way I always think about it is, well I have
- a cosine and a sin.
- The cosine is squared, so all these crazy things are
- happening to it, so it seems like I could use substitution
- or the reverse chain rules some out here.
- And what was the chain rule?
- The chain rule said if I took the derivative of f of g of x
- that this is equal to f prime of g of x times g prime of x.
- That might completely confuse you, but I just wrote that here
- because we could say, well what if g of x is cosine of x. f
- prime of g of x is the cosine of x squared, and then the
- derivative of g of x or the derivative of cosine
- of x is sin of x.
- Well it's actually minus sin of x, and we have a minus sin
- here, so that works out well too.
- If this confuse you, ignore it.
- Well essentially we're just going to do the same thing,
- but we're going to do it with substitution.
- So let me do it with substitution.
- Let me a erase this if this confuses people.
- I want to do it however it is least confusing to you.
- OK let me erase that.
- Actually, let me do it with substitution, just because the
- way I was just doing is kind of my shortcut back in the
- day when I was a mathlete.
- But it's good to be able to do it with substitution, helps you
- from making careless mistakes.
- So let me rewrite this first of all.
- This is the same thing as the integral from pi over 2 to
- pi of cosine squared of x.
- Actually let me write that as cosine of x squared.
- Same thing, right.
- Times minus sin of x dx.
- And now it should be clearer to you.
- What's the derivative of cosine of x?
- It's minus sin of x.
- So I have a function and it's being squared, and I have its
- derivative, so I can figure out its antiderivative by using
- substitution or the reverse chain rule.
- So let's make a substitution.
- u is equal to cosine of x.
- How did I know to substitute u is equal to cosine of x?
- Well because I say, well the derivative of this
- function is here.
- So when I find du, this whole thing is going to end up
- being du, and let me show that to you.
- So what is du/dx?
- du/dx is equal to minus sin of x.
- That hopefully we've learned already.
- So what is du?
- If we multiply both sides by the differential d of x get du
- is equal to minus sin of x dx.
- So if we look at the original equation, this right here we
- just showed is equal to du, and this right here is what?
- Cosine of x is u.
- That was our original substitution.
- So we have u squared.
- So now let's take the integral.
- And I will arbitrarily switch colors.
- And now this is a very important thing.
- If you're going to do substitution, if we're going to
- say u is equal to cosine of x, we're going to have to actually
- make this substitution on the boundaries.
- Or we could do the substitution and reverse the substitution
- and then evaluate the boundaries, but let's do that.
- So if this is going from x is equal to pi over 2 to pi,
- what is u going from?
- Well when x is equal to pi over 2, u is equal
- to cosine of pi over 2.
- Because u is just cosine of x.
- And then when x is pi, i is going to be cosine of pi.
- And now the fun part.
- Cosine of x squared.
- Well that's just the same thing is u squared.
- And minus sin of x dx, that's the same thing
- is du-- did that here.
- This is pretty straightforward and I'm just going
- to rewrite it.
- What's cosine of pi?
- Cosine of pi is minus 1.
- Cosine of pi over 2, well that's a 0.
- So we have the integral from u is equal to 0 to u is equal to
- negative 1 of u squared du.
- And now this seems like a simple problem.
- So this is equal to the antiderivative of u
- squared, which is fairly straightforward.
- u cubed over 3.
- You could just take the derivative of this,
- you get this.
- All I did is I increased this exponent to get the third, and
- I divided by that exponent.
- Now we're going to have to evaluate it at minus 1 and
- subtract from that, it evaluated at 0.
- So this is equal to minus 1 to the third over 3 minus
- 0 over 3, and so this is equal to minus 1/3.
- And we are done.
- And if we look at this area from our original graph, what
- we just solved, as we said the area of the curve between
- here and here is minus 1/3.
- Or if we wanted the absolute area because you can't really
- have a negative area, it's 1/3 but we know it's negative
- because this curve is below the x-axis here.
- And that looks about right, that looks about 1/3.
- I mean if this cube right here is 1, then that
- looks about 1/3.
- The intuition all works out at least.
- So hopefully you found that vaguely useful.
- Actually let me-- since we have a little bit of time.
- Hopefully you understood this, and if you did, don't worry
- about what I'm going to do now, but I want to show you how I
- tend to do it where I just think of it is the
- reverse chain rule.
- It's a little bit quicker sometimes.
- But it's really the same thing as what we just
- did with substitution.
- So if we erase all of this.
- And so if we have this integral right here, what I do is I say,
- well I have cosine of x squared.
- I have cosine of x squared, and then I have minus sin of x.
- This is the derivative of this.
- Since the derivative is here, I can just treat this whole
- thing like an x term.
- So this is the same thing.
- So the antiderivative is cosine of x to the third over 3, and I
- evaluate it at pi and pi over 2.
- And remember, how did I do this?
- What allowed me to treat this cosine of x just like an x or
- like a u when I did it with the substitution?
- Well I had its derivative sitting right here,
- minus sin of x.
- So that's what gave me the license to just take the
- antiderivative, pretend like this cosine of x is just an x,
- is just a u, you could even say, and just take it's
- exponent, raise it by 1 and divide it by 3 and then
- evaluate it from pi to pi over 2.
- So this is equal to cosine of pi cubed over 3 minus cosine
- of pi over 2 cubed over 3.
- This is minus 1 to the third, so this is equal to minus
- 1/3 minus this is 0.
- So we get the same answer.
- I just wanted to show you that.
- It's exactly the same with substitution.
- It's just I didn't formally do the substitution, but it's
- the exact same thing.
- Anyway, hope you found that helpful.
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