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Trig and u substitution together (part 2)

More of all of the substitution! Created by Sal Khan.

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  • blobby green style avatar for user Justin  Moreno
    Is there anything wrong with just plugging in cosθ for all the u's and then plugging in arcsin(x/3) for all the θ's?
    (54 votes)
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  • starky ultimate style avatar for user Peter Kapeel
    so, in general, when given an integral where 2 functions are multiplied or divided one should:
    1) Check to see if Trig substitution is applicable, if not
    2) Check to see if U Substitution is applicable, if not
    3) Integrate by parts

    would this be the optimal algorithm for calculating integrals of this format or is there a better way?
    (15 votes)
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  • leaf green style avatar for user Chris
    Where did the -243 go? You copied it as 243 in the following line.
    (4 votes)
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  • piceratops ultimate style avatar for user Kartikeye
    How did the 1/3 become 1/9 when it was put in the radical?
    (4 votes)
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  • leafers ultimate style avatar for user Muhannad AlAyoubi
    We can use the Pythagorean identity, right? Here's what I've done:
    cos θ = u, sin θ = x/3
    cos^2 (θ) + sin^2 (θ) = 1
    u^2 + (x^2)/9 = 1
    u^2 = 1 – (x^2)/9
    u = sqrt (1 – (x^2)/9)
    which is the same as sir Sal's result.
    (5 votes)
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  • blobby green style avatar for user Josh
    At , Sal says that u=cos(theta)=sqrt(1-sin^2(theta)). I thought that sqrt(1-sin^2(theta)) = cos^2(theta)
    (2 votes)
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  • duskpin ultimate style avatar for user staceyrivet
    you can just to u-sub and avoid trig sub. change x^3 to x and x^2 where x^2 = 9-u...and du/dx = -2x. It saved my sanity.
    (2 votes)
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  • leaf green style avatar for user miles marcuson
    Couldn't you just solve for θ by taking the inverse sine of both sides of (x/3)=sin(θ) to get θ=arcsin(x/3) and then u=cos(arcsin(x/3))?
    (1 vote)
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  • purple pi pink style avatar for user Sachin
    At , would it be okay to say that cos θ = sin 90-θ? So if x/3 = sin θ, then u = 90 - x/3, right?
    (2 votes)
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    • leaf grey style avatar for user Acalc79
      Where did the part "cos θ = 90 - sin θ" come from? I think it is incorrect, actually I'm pretty sure, because the range of cos and sin is the set [-1, 1], and your equation implies that either sin or cos have values way beyond that interval.
      (0 votes)
  • blobby green style avatar for user Aditya Vasudevan
    Wouldn't it have been simpler to not take 1/3 inside the root while substituting for cos? Because in that case, you can distribute 243, which is 3^5, which cancels out neatly with quite a lot of stuff. It gives the answer as {(9-x^2)^(5/2)}/5 - 3(9-x^2)^(3/2) + c. Is there a specific reason, why the answer has been expressed as in the video?
    (1 vote)
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Video transcript

In the last video, in order to evaluate this indefinite integral, we first made the substitution that x is equal to 3 sine theta. And then this got us to an integral of this form. Then we were able to break up these sines and cosines and use a little bit of our trig identities. To get it into the form where we could do u substitution, we did another substitution where we said that u is equal to cosine of theta. And then finally, we were able to get it into a form using that second round of substitution. And this time, it was u substitution. We were able to get it into a form that we could actually take the antiderivative. And we got the final answer here in terms of u. But now we've got to go and undo everything. We have to undo the substitutions. So the last substitution we had done-- we're now going to go in reverse order-- was that u was equal to cosine theta. So you might just want to substitute u with cosine theta here. But then we're going to have everything in terms of cosine theta, which still doesn't get us to x. So the ideal is if I can somehow express u in terms of x. So let's think how we can do it. We know that u is equal to cosine theta. We know the relationship between x and theta is right over here. x is equal to 3 sine theta. So let's write that over here. So we know that x-- let me write it over here-- we know that x is equal to 3 sine theta. So if we could somehow write cosine-- let me rewrite this a little bit differently. Or we could also say that x over 3 is equal to sine theta. I just divided both sides by 3. So if we could somehow re-express this in terms of sine theta, then we can replace all the sine thetas with x over 3's, and we are done. So how can we do that? And I'll actually show you two techniques for doing it. So the first one is to make the realization, OK, u is equal to cosine of theta. If I want to write this in terms of sine of theta, I can just say that this is equal to-- straight up, this is the most fundamental trigonometric identity. Cosine theta is the square root of 1 minus sine squared theta. And we see sine of theta is equal to x over 3. So this is the square root of 1 minus x over 3 squared. So this is u in terms of x. So everywhere we see a u up here we can replace it with this expression. And we are essentially done. We would have written this in terms of x. Now, there's another technique you might sometimes see in a calculus class where someone says, OK, we know that u is equal to cosine theta. We know this relationship. How can we express u in terms of x? And we'll say, let's draw a right triangle. They'll draw a right triangle like this. They'll draw a right triangle, and they'll say, OK, look, sine of theta is x over 3. So if we say that this is theta right over here, sine of theta is the same thing as opposite over hypotenuse. Opposite over hypotenuse is equal to x over 3. So let's say that this is x and then this right over here is 3. Then the sine of theta will be x over 3. So we look at that first substitution right over here. But in order to figure out what u is in terms of x, we need to figure out what cosine of theta is. Well, cosine is adjacent over hypotenuse. So we have to figure out what this adjacent side is. Well, we can just use the Pythagorean theorem for that. Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared, which is 9, minus the other side squared, minus x squared. So from this, we fully solved the right triangle in terms of x. We can realize that cosine of theta is going to be equal to the adjacent side, square root of 9 minus x squared, over the hypotenuse, over 3, which is the same thing as 1/3 times the square root of 9 minus x squared, which is the same thing if we square 1/3 and put it into the radical. So we're essentially going to take the square-- 1/3 is the same thing as the square root of 1/9. So can rewrite this as the square root of 1/9 times 9 minus x squared. Essentially, we just brought the 1/3 third into the radical. Now it's 1/9. And so now this is going to be the same thing as the square root of 1 minus x squared over 9, which is exactly this thing right over here. x squared over 9 is the same thing as x over 3 squared. So either way, you get the same result. I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do the substitution to be a little bit more straightforward. But now we can just substitute into the original thing. So either of these-- I can write it as either way-- this thing right over here, this is the same thing as 1 minus x squared over 9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just substitute it with this thing. So our final answer in terms of x is going to be equal to 243 times u to the fifth, this to the fifth power over 5. This to the fifth power is 1 minus x squared over 9. It was to the 1/2, but if we raise that to the fifth power, it's now going to be to the 5/2 power over 5 minus this to the third power, 1 minus x squared over 9 to the 3/2 raising this to the third power-- that's this right over here-- over 3, and then all of that plus c. And we're done. It's messy, but using first trig substitution then u substitution, or trig substitution then rearranging using a couple of our techniques for manipulating these powers of trig functions, we're able to get into a form where we could use u substitution, and then we were able to unwind all the substitutions and actually evaluate the indefinite integral.