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𝘶-substitution: double substitution

Finding the indefinite integral of cos(5x)/e^[sin(5x)]. To do that, we need to perform 𝘶-substitution twice. Created by Sal Khan.

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• Okay, so I tried solving the problem my own way and the answer I got puzzles me.

I did the same as Sal up until where I just decided to take the indefinite integral of 1/e^u, and this gave me (1/5)*(ln|e^u| + C). By the properties of the natural logarithm, wouldn't this simplify to (1/5)*(u + C)? When substituting back the expression instead of u, I don't get the same result as Sal did. I assume the absolute value messes up my reasoning somehow, and my question is: Why? I can't think of any possible situation where |e^u| would not be equal to (e^u).
• The indefinite integral of 1/e^u isn't ln |e^u| + C, it's -1/e^u + C.

Anything of the form f´(x)/f(x) integrates to ln |f(x)| + C, 1/e^u clearly isn't of that form.
• Can't we just make a general property that integration of e^-x is -e^x?
• The integral of e^(-x) is DEFINITELY not -e^x. The integral of e^(-x) is -e^(-x). The x in the exponent is negative as well.

As for making "general properties," in these integral things, usually they only make general properties out of the basic things you need. Since we already know that the integral of e^x is e^x, we can figure out the integral of e^(-x) using skills we already know.

In short, it's not made into a general property since you already are able to solve it using other general properties. That being said, if you want to memorize it to be able to use it more quickly, feel free to do so.
• Why substitute again what's wrong with e^-u ??
• You could integrate e^-u in your head, but it may not be clear to some people what the antiderivative of e^-u is. Therefore, we must use u-substitution to change the integral into one we can solve, namely integral of e^u, which is very simple.
(1 vote)
• How do you know when to do u substitution twice?
• I think it's just pretty much that if you have a problem that's seems to be workable with u-sub, but you don't quite "get there" on the first go. Then you do it again. :)

My guess is that most of the time you will only notice that you need to do it double towards the end of the problem.

Hope that helps!
• Why e^-u at .I dont get it!
• Basic property of powers in a fraction: those on the bottom are negatives of those on the top. 1/x = x^(-1) Therefore, 1/e^u = e^(-u) to get rid of the fraction.
• I don't quite get why Sal had to do the u-sub twice. 1/5 int e^-u du would be simply -1/5 e^-u + c?
• Yes, you are correct. You don't have to do that second substitution, but you can do it. When you learn how to integrate by the reverse chain rule (if you haven't already), you can do this problem without any substitution.
• Why d(sin5x)/dx = 5cos5x ? Please explain.
(1 vote)
• You have to use the chain rule here
d/dx (sin x) = cos x so if there is a function inside instead of just an x you do
d/dx (sin (f(x))) = cos (f(x))*d/dx(f(x))
in other words find the derivative of sin then multiply by the derivative of the inside function so...
d/dx (sin (5x)) = cos(5x)*d/dx(5x)= cos(5x)*5 = 5cos(5x)
• Instead of multiplying the whole integral by 5/5, could you not just make the substitution 1/5 du = cos(5x) dx ?
• You could make the substitution that way as well. It is merely a stylistic preference.