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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 5

Lesson 3: Reverse chain rule

# Reverse chain rule introduction

What is the "reverse chain rule", and why it does the same thing as u-substitution.

## Want to join the conversation?

• In what circumstances would it be necessary to resort to u substitution? •   Good Question.
U sub is a method for algebraically simplifying the form of a function so that its anti-derivative can be more easily recognized. This method is intimately related to the chain rule for differentiation, which when applied to anti-derivatives is sometimes called the reverse chain rule. If you can do all the algebraic manipulation and simplification in your head, then you don't actually need to explicitly write out the the u substitution you performed, though you are actually doing it, that is, recognizing some f and f' within the integrand. You will find over time, as you do more and more problems, you will be able to do more and more in your head, and that is fine. Sometimes it is helpful to write out the u sub so as just to avoid careless algebraic errors, however, everyone is different with respect to their personal comfort zone and preferences when choosing what to write out when simplifying expressions and what not. U sub is just one simplification technique of many in our toolbox.

On a personal note, I have found that the exercises here at the Khan academy don't spend as much time converting seemingly intractable anti-derivatives into an integrable form as when I was in university - it made integration much more of an art form than differentiation, which was pretty much cook-book math: recognize the form, apply the appropriate recipe. We used to call the art of integration "Algebra and Innovation."
I hope this answer was satisfactory.
Keep Studying!
• I am having a lot of trouble identifying g(x). I understand that sinx is f(x) and cosx is f'(x), so is the ^2 g' in this case? • Isn't this basically just u-substitution? Is there a particular difference or no? • integration of e^atanx? • Why exactly does the cos x disappear? I understand how he gets ((sin x)^3)/3 but where'd the cosine go? • When doing the µ-substitution we replace the integrand with new symbols. Symbols that are in the first integrand are sinx, cosx and dx. These all depend on x, so we need to make them all depend on µ. Here is how the substitution was made:

µ=sin(x)
dµ=cos(x)dx <=> [dx=dµ/cos(x)]

Now dx in the first integrand can be replaced with dµ/cos(x), so you get the new integrand:

µ^2*cosx*(dµ/cosx)=µ^2dµ

If we calculate this integrand µ^2dµ we get the same solution of (µ^3)/3+C, which from the substitution equals [sin(x)^3]/3+C. (dont forget the constant at the end!)
• Isn't the integral of g'(f(x)) = (g(f(x))/(f'(x))+C? E.g. ∫e^(2x) = (e^(2x))/2. • This trick only works when f'(x) is a constant, which means f(x) is of the form ax + b. So it's true that ∫e^(2x)dx = (e^(2x))/2, and ∫e^(ax + b)dx = (e^(ax + b))/a.

But it is NOT true that ∫e^(x²)dx = (e^(x²))/(2x). (FALSE!) You can see that if you go to check your antiderivative by taking its derivative, if f'(x) is not a constant then you need the quotient rule and it won't be correct.
• Wouldn't Cosx be changed to -Sinx since we're integrating here? • So in u-substitution the du is the derivative of f(x) so when we take the integral of it we have the original function (according the the Fundamental Theorem) and so the du part doesn't necessarily disappear but instead it is 'inside the composite function g(f(x)). Is this right? • when we use the u-subtitution technique we have to calculate the integral in a "new interval" because of the substitution : I=[a.b] "becomes" J=[g(a) . g(b)].
What about the reverse chain rule ? Are we calculating in I or in J ? thx ^^ • I am struggling with the integral x(x+6)^1/2. When you attempt to derive the inner function, you get x (outer function) is = 1 (derivative of the inner function)

I see a pattern with similar problems, but can you use the reverse chain method?
(1 vote) • Ah, the "x" in your function is NOT an outer function. In fact, outer function is something appeared in composite function, ie in the form g(f(x)). Here "x" is just being multiplied with the other thing. So this can't be solved by reverse chain rule.

So just use u-substitution. Denote u=x+6, and du=dx.
Int ( x(x+6)^(1/2) dx ) = Int ( (u-6)u^(1/2) du ).
Now distribute the u^(1/2) to u-6, which gives Int ( u^(3/2) - 6u^(1/2) du )
And then use Reverse Power Rule, which gives 2/5u^(5/2) - 6(2/3)u^(3/2) +C, the exact same as what you've posted.

As Sal mentioned, Reverse Chain Rule is just u-substitution. Indeed it is faster but it is likely to get the wrong answer.

Hope this helps!