Basic Integration Problems
In this video, I want to talk about some basic integration problems. And by "basic", I mean You can either do them either by doing a little bit of algebra to your integrand, or you can just use a trig identity. No u-substitution, no partial fractions nothing any crazier than that. So, suppose we want to integrate (1+2x-4x^3) and all the problems I'm gonna do right now are going to be indefinite integrals.
The idea when there's pluses or minuses in between your terms You can basically integrate them a piece at a time. Typically, you won't really write it like this Most people will go ahead and just start integrating. That's okay, just for our first example we'll rewrite it. Basically, it's kind of like you're distributing the integral sign to each piece.
The anti-derivative of 1 is just x. Just like when you take derivatives, the constant gets multiplied through at the end. When you find anti-derivatives, the constant also gets multiplied through at the end. So, the anti-derivative of x^1-- remember you add one to the power and also divide by that new power-- minus 4x^4 over 4 Remember with indefinite integrals, you put this "plus C" on there So I'm going to simplify it; my 2's will cancel I'll have x+x^2 My 4's will cancel x^4 + c and there is my indefinite integral for this first problem.
Hopefully nothing too tricky. So obviously, you need to know the basic formulas that go with these, but after that, it's just a matter of applying those formulas. Suppose we have 3x-2 divided by the square root of x The first thing I usually do on these problems is the same thing as if I was going to take a derivative. Any radicals I'll usually go ahead and rewrite those as powers So my square root of x I'll rewrite as x^1/2 At first sight, you may think, "Well, I'll just take the anti-derivative of each piece" but as a rule of thumb, you use the same idea you would for taking derivatives.
If you're going to take the derivative of 3x-2 over x^1/2, you wouldn't take the derivative of each piece individually. You'd have to use the more complicated quotient rule for taking derivatives. Well, for finding anti-derivatives, it's the same way. You can't just find anti-derivatives of each piece.
But what you CAN do on this one is a little bit of algebra. You could rewrite it as 3x over x^1/2 minus 2 over x^1/2 At this point, I'm going to simplify things down. So 3x divided by x^1/2 If I simplify x^1 over x^1/2, I'm left with x^1/2 in the numerator minus 2 times x If I bring the x^1/2 to the numerator, it will become x^-1/2 And now this is in a suitable form for me to be able to integrate this. So, again, the 3x comes along for the ride If I add 1 to the 1/2, I'll get 3/2 And then I'll have to divide by 3/2 I'll add 1 to -1/2 and get 1/2 Again, I have to divide by 1/2 plus C.
If we simplify this, dividing by 3/2 is the same thing as multiplying by 2/3, so the 3's will actually cancel out, And I'll have 2x^3/2 Same thing; Dividing by 1/2 is the same thing as multiplying by 2 so minus 4x^1/2 plus C. Again, there is your anti-derivative. If it doesn't look like something very basic and fundamental, then usually the first thing I try to do is to break it up using some algebra. In terms of all the integration techniques you'll learn, usually tweaking things with a little bit of algebra will be the most straightforward way to go about things.
Let's do another one here. Let's throw some trig into this one. Suppose we have sin + sintan^2 and all of that being divided by sec^2. Maybe you could try the same trick as in the last problem -- maybe that would work, maybe not.
But the trick to this one is recognizing there's a sin on top -- you can actually factor that out You'll have sin then (1+tan^2) all being divided by sec^ The trick to doing a lot of problems involving trig is simply knowing your trig identities. So, recall that sin^2 + cos^2 equals 1. This is one of the fundamental trig identities. If you divide everything by cos^2, Well, sin/cos is tan, but I have two of those, so I'm going to get tan^2 cos^2/cos^2 is 1 1/cos is sec, but again, it's being squared, so I'm going to get sec^2 Well, that's what I have right here in the top of my fraction: 1+tan^2 So, if you don't know your trig identities, you're going to run into a ton of problems with integration, because a lot of times, that's simply going to be the trick is recognizing a trig identity.
And if you don't see it, it may be really hard -- if not impossible to do. So 1+tan^2, again, is the same thing as sec^2 I'll divide that by sec^2 Well, I can just cancel out my sec^2 so that, really, all I'm integrating now is sin. And the anti-derivative of sin not quite cos, because remember, the derivative of cos is -sin, so we need to put an extra minus sign in there, and then we'll tag on our plus c. So for something that looked like a pretty complicated integral at first, we end up with a relatively simple anti-derivative.
Let's maybe just do one more here, since we're going to run out of time. So suppose here you have (x+1) times (x^2+3) Well, the same idea on this one; You can't just find the anti-derivative of each piece. The first thing you have to do on this one is some algebra, and to simply multiply it out. So x times x^2 is x^3 x times 3 is positive 3x 1 times x^2 is positive x^2 1 times 3 is positive 3 If there's any like terms, combine them; Obviously, there's none in here so we can't really do much simplification on the integrand.
And now we'll just integrate this thing So the anti-derivative of x^3 is x^4 over 4 The anti-derivative of x^1 is x^2 over 2 Again, we're just using this property of adding 1 and dividing by the new number. We'll get x^3 over 3 The anti-derivative of 3 is simply 3x and then we'll just tack on our plus c. And again, there's your anti-derivative of the original problem. Again, I think this is a good rule to keep in the back of your mind.
If it's not something very basic, just think to yourself, "Can I either do some algebra or trig identities?" There's definitely other techniques for integrating things, and integration in general can become pretty tedious and pretty difficult, so usually, doing some algebra or some trig identities is probably going to be the easiest thing to do if it does work and if it doesn't work, you can quickly rule it out without wasting too much time. Hope these videos helped! Definitely feel free to jump over to my website for a ton of other calculus videos and other topics as well! :) For a ton of other calculus videos and other topics as well !:).
The idea when there's pluses or minuses in between your terms You can basically integrate them a piece at a time. Typically, you won't really write it like this Most people will go ahead and just start integrating. That's okay, just for our first example we'll rewrite it. Basically, it's kind of like you're distributing the integral sign to each piece.
The anti-derivative of 1 is just x. Just like when you take derivatives, the constant gets multiplied through at the end. When you find anti-derivatives, the constant also gets multiplied through at the end. So, the anti-derivative of x^1-- remember you add one to the power and also divide by that new power-- minus 4x^4 over 4 Remember with indefinite integrals, you put this "plus C" on there So I'm going to simplify it; my 2's will cancel I'll have x+x^2 My 4's will cancel x^4 + c and there is my indefinite integral for this first problem.
Hopefully nothing too tricky. So obviously, you need to know the basic formulas that go with these, but after that, it's just a matter of applying those formulas. Suppose we have 3x-2 divided by the square root of x The first thing I usually do on these problems is the same thing as if I was going to take a derivative. Any radicals I'll usually go ahead and rewrite those as powers So my square root of x I'll rewrite as x^1/2 At first sight, you may think, "Well, I'll just take the anti-derivative of each piece" but as a rule of thumb, you use the same idea you would for taking derivatives.
If you're going to take the derivative of 3x-2 over x^1/2, you wouldn't take the derivative of each piece individually. You'd have to use the more complicated quotient rule for taking derivatives. Well, for finding anti-derivatives, it's the same way. You can't just find anti-derivatives of each piece.
But what you CAN do on this one is a little bit of algebra. You could rewrite it as 3x over x^1/2 minus 2 over x^1/2 At this point, I'm going to simplify things down. So 3x divided by x^1/2 If I simplify x^1 over x^1/2, I'm left with x^1/2 in the numerator minus 2 times x If I bring the x^1/2 to the numerator, it will become x^-1/2 And now this is in a suitable form for me to be able to integrate this. So, again, the 3x comes along for the ride If I add 1 to the 1/2, I'll get 3/2 And then I'll have to divide by 3/2 I'll add 1 to -1/2 and get 1/2 Again, I have to divide by 1/2 plus C.
If we simplify this, dividing by 3/2 is the same thing as multiplying by 2/3, so the 3's will actually cancel out, And I'll have 2x^3/2 Same thing; Dividing by 1/2 is the same thing as multiplying by 2 so minus 4x^1/2 plus C. Again, there is your anti-derivative. If it doesn't look like something very basic and fundamental, then usually the first thing I try to do is to break it up using some algebra. In terms of all the integration techniques you'll learn, usually tweaking things with a little bit of algebra will be the most straightforward way to go about things.
Let's do another one here. Let's throw some trig into this one. Suppose we have sin + sintan^2 and all of that being divided by sec^2. Maybe you could try the same trick as in the last problem -- maybe that would work, maybe not.
But the trick to this one is recognizing there's a sin on top -- you can actually factor that out You'll have sin then (1+tan^2) all being divided by sec^ The trick to doing a lot of problems involving trig is simply knowing your trig identities. So, recall that sin^2 + cos^2 equals 1. This is one of the fundamental trig identities. If you divide everything by cos^2, Well, sin/cos is tan, but I have two of those, so I'm going to get tan^2 cos^2/cos^2 is 1 1/cos is sec, but again, it's being squared, so I'm going to get sec^2 Well, that's what I have right here in the top of my fraction: 1+tan^2 So, if you don't know your trig identities, you're going to run into a ton of problems with integration, because a lot of times, that's simply going to be the trick is recognizing a trig identity.
And if you don't see it, it may be really hard -- if not impossible to do. So 1+tan^2, again, is the same thing as sec^2 I'll divide that by sec^2 Well, I can just cancel out my sec^2 so that, really, all I'm integrating now is sin. And the anti-derivative of sin not quite cos, because remember, the derivative of cos is -sin, so we need to put an extra minus sign in there, and then we'll tag on our plus c. So for something that looked like a pretty complicated integral at first, we end up with a relatively simple anti-derivative.
Let's maybe just do one more here, since we're going to run out of time. So suppose here you have (x+1) times (x^2+3) Well, the same idea on this one; You can't just find the anti-derivative of each piece. The first thing you have to do on this one is some algebra, and to simply multiply it out. So x times x^2 is x^3 x times 3 is positive 3x 1 times x^2 is positive x^2 1 times 3 is positive 3 If there's any like terms, combine them; Obviously, there's none in here so we can't really do much simplification on the integrand.
And now we'll just integrate this thing So the anti-derivative of x^3 is x^4 over 4 The anti-derivative of x^1 is x^2 over 2 Again, we're just using this property of adding 1 and dividing by the new number. We'll get x^3 over 3 The anti-derivative of 3 is simply 3x and then we'll just tack on our plus c. And again, there's your anti-derivative of the original problem. Again, I think this is a good rule to keep in the back of your mind.
If it's not something very basic, just think to yourself, "Can I either do some algebra or trig identities?" There's definitely other techniques for integrating things, and integration in general can become pretty tedious and pretty difficult, so usually, doing some algebra or some trig identities is probably going to be the easiest thing to do if it does work and if it doesn't work, you can quickly rule it out without wasting too much time. Hope these videos helped! Definitely feel free to jump over to my website for a ton of other calculus videos and other topics as well! :) For a ton of other calculus videos and other topics as well !:).
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