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# 𝘶-substitution: multiplying by a constant

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
Manipulating the expression to make u-substitution a little more obvious. Created by Sal Khan.

## Want to join the conversation?

• @ you say "even though it wasn't completely obvious we see that u-sub is applicable. How can we tell when it IS NOT applicable? Do you have a video for "u-sub, to use or not to use?". Thanks for a great site.
Dallas
• If you end up with an equation that you find difficult to solve otherwise, then you could try u-sub. Like in this example if you feel that Sqrt(7x+9) is difficult to integrate u-sub could be an easier way to solve it.
But, u-sub never is the only solution! If there is a way to solve it by u-sub, then there is a way without it. After all, all you do is renaming things (by saying that u = 7x+9 for example) and rearranging numbers.

So while it is very good to realise when u-sub can be done it is also good to realise that you dont always have to.
And the same goes the other way around. U-sub is a way to try to make the equation clearer for yourself. So naturally you should not use it when if the equation already is easy to solve. Or if the equation in its current form naturally lends itself to some rule or standard integral that you know. The more rules/standard integrals you know the easier it becomes to know when the best way to solve an equation is with or without an u-sub. So learning when to u-sub is not about practicing u-sub only but learning all other things as well.
• Sorry if this seems silly, but I still do not quite understand where the 1/7 comes from at instead of just 7?
• Unfortunately, we can't simply make up a constant to bring in. To remedy this, he effectively multiplies the primary equation by 1, which we can do, since 1 multiplied by any number is itself. 1/7*7=7/7=1, so by using this method, he is able to bring in a constant without changing the primary equation to something it is not.
• at , when that 1/7 is distributed to the rest of the problem, wouldn't the result end up with a 1/7 C?
• Any constant multiplied by a scalar is still just some constant. Right after that Sal talks about how you could call them C_1 and C_2 to show that they are different constants. However, this precision is not normally required and is often left out entirely since it doesn't change the result. If you wanted to be thorough, you would call them C_1 and C_2 and also state that C_1 = 1/7 C_2.
• So is it typically a good strategy to choose "u" as something that either is (1) in the denominator, (2) in parenthesis, or (3) under a radical? I'm just looking for good strategies to choosing good "u" candidates. Thank you.
• Whatever you choose as u, you MUST have the derivative of that thing you call u being multiplied by whatever standard integral function you wish to use.
All of the standard integral forms you will learn amount to:
∫ f(u) du where du is NOT just some notation, it needs to be treated as the derivative of whatever you call u.

So, an effective strategy is to break the function into two pieces that are multiplied by each other. The first piece will be something you can use u substitution for to get the integral into a form you know how to integrate. The second piece will be the derivative of whatever you called u. If you cannot do that, you cannot use u substitution.

An example I recently gave to another post. Suppose you had:
∫ cos(x²) dx you might think make u=x² and this becomes
∫cos (u) du which is easy to integrate. But this is WRONG.
To use u=x² you must also have du = 2x dx. There is no 2x in the integral, so you cannot use that substitution. You would need:
∫ 2x cos (x²) dx you have u=x² and du = 2x dx and that gives you:
∫ cos (u) du = sin (u) + C = sin (x²) + C
It turns out, though it looks simpler, ∫ cos(x²) dx cannot be integrated by any means taught in introductory integral calculus courses, but is a very advanced level problem.

So, remember, the dx or du is not just some notation you can tack on.
• What happens to du when you take the integral of du? It just disappears? What happens when you take the derivative of dx? does it disappear too? WHY R THINGS DISAPPEARING?
• The derivative of 'dx' on its own is zero, because you're deriving 1 dx, and the derivative of a constant is zero. You can however derive f'(x)dx, and you get the second derivative of your function f(x), or f"(x)dx. 'dx' means "a little change in the x-direction", so when integrating, you're multiplying the function by these minute changes. Remember Riemann sums? If you write integrals in terms of the sums, it looks like this: sigma f(x) delta(x); in other words, you're summing the area of all of the little rectangles created by your function taken over small changes in x, or delta(x). If these changes are truly minuscule, then delta(x) is transformed into 'dx', and the sigma, or sum, into the integral. I hope this helps!
• Im not seeing how U-Sub is basically anti-chain rule. Could someone attempt to make this more clear? Thank you.
• Watch the first video on u-substitution (https://www.khanacademy.org/math/calculus/integral-calculus/u_substitution/v/u-substitution). The integral he has there is
∫ (3x² + 2x) e^(x³ + x²) dx.
This integral evaluates to e^(x³ + x²) + C.

You will notice that if you take the derivative of e^(x³ + x²) using the chain rule you get
d/dx e^(x³ + x²) = (x³ + x²)′ e^(x³ + x²) = (3x² + 2x) e^(x³ + x²),
which is in fact the expression whose antiderivative you were trying to find in the first place.

Therefore, u-substitution can in such cases be used to unravel the chain rule.
• If I don't use U-substitution will I get the same answer?
• U-substitution is never actually required. I often don't use it. However, not using it requires keeping track of a lot of different things. U-sub greatly simplifies keeping track of things so you don't overlook something important.

However, if you are able to keep track of everything and do it all without the u-sub, then go for it.
• Can you use indefinite integrals to find out definite integrals?
Sort of like how you can use derivatives to figure out the tangent line of a point?
• Yes look at the fundamental theorem of calculus. Definite and indefinite integrals are essentially the same thing only one of them you are applying the first part of the fundamental theorem of calculus.