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# 𝘶-substitution: defining 𝘶 (more examples)

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
A common challenge when performing 𝘶-substitution is to realize which part should be our 𝘶.

## Want to join the conversation?

• What would the answer be for both problems after you put your answer back in terms of x? Do you have to take the antiderivative of u^10 and then plug ln(x) back in for u and plug 1/x back in for du?
• Greetings
Here are the solutions to the two problems:

1) ∫ ((ln (x))^10)/x) dx= 1/11 ln (x)^11 + C

2) ∫ tan(x) dx = -ln |cos(x)| + C

I hope this helps!
• huh? but my teacher gave me a list if integral formulas and said that integral of tanx =
log l sec(x)) l + c

i am not sure as to what i should believe... does Sal's answer somehow simplify down into this? and if so, how?
• They are equivalent! Following through with Sal's solution, we arrive at:
-log |cos 𝑥| + 𝐶
But recall that 𝑎log 𝑏 = log(𝑏ª). Hence:
-log |cos 𝑥| + 𝐶 = log |(cos 𝑥)⁻¹| + 𝐶
The reciprocal of cosine is secant, hence:
-log |cos 𝑥| + 𝐶 = log |(cos 𝑥)⁻¹| + 𝐶 = log |sec 𝑥| + 𝐶
As desired. Comment if you have questions!
• for the tan x question, can I instead plug in u= sin x and u' = cos x?
• Actually No, because the dx term wouldn't have been able to be replaced. here, let's see how it works out.

u = sin x
du/dx = cos(x)
du = cos(x) dx

Meanwhile in the problem you start with sin(x)/cos(x) dx. if you replace sin(x) with u you have:

u/cos(x) dx or u (1/cos(x))dx but du = cos(x) dx so you can't sub that out for du. Let me know if that doesn't make sense.
• i tried to solve the second example in other form,

int(tan(x))dx
1/(sec(x))^2int(tan(x)(sec(x))^2 dx

u= tg(x)
du=(sec(x))^2dx

1/(sec(x))^2 int(u du)

1/((sec(x))^2 u^2/2 +c

(tg(x))^2/2(sec(x))^2 +c

(sen(x))^2/2 +c

did i made a mistake ??
• We can't move functions outside of the integral.

, ∫𝑓(𝑥)𝑑𝑥 = ∫𝑓(𝑥)𝑔(𝑥)∕𝑔(𝑥)𝑑𝑥 ≠ 1∕𝑔(𝑥)∙ ∫𝑓(𝑥)𝑔(𝑥)𝑑𝑥
• I wanna know if it is possible to simplify ∫ln(x)^10* dx to be 10∫ln(x) first before substituting. Because then the final answer would be 10 * u^2/2 = [5ln(x)^2 + C] instead of ln(x)^11/11 + C
• Look at the brackets carefully. It is (ln(x))^10 NOT ln(x^10). Theres a differene, we're raising whatever ln x is to the power 10 instead of raising x to the power 10 and then finding the resulting ln. Hence we cannot use that log property
(1 vote)
• Hey, Sal. How would you solve an integral if the integrand doesn't contain the derivative of the function, given that the derivative of said function is not a constant, but another function with variables? My teacher taught us that you can only multiply by constants to create the derivative.

Ex: Integral of 3 (sin x)^-2 dx

My first instinct was to set u = sin x. However, that would result in du = cos x dx, and you can't create that derivative by only introducing constants. My next thought was to set u equal to something else, but I don't know what.

Follow-up question: How do you know what to set u equal to in a question like this?
• What happens when the expression has a constant .... like, how to do this one indefinite integral of x/(x^2 + a^2)^3/2 ....??
Plz solve in detailed manner,
Thanx.
• Take u = a^2 + x^2.
du = 2x dx.
Write the numerator in terms of du, the denominator in terms of u, and apply the reverse power rule.
• Isn't the antiderivative of (1/x) actually ln(|x|) - with the absolute value?
(1 vote)
• The ln(x) must have an absolute value sign on it! ln(|x|)
• If we were integrating 1∕𝑥, then we would have to assume that 𝑥 can be negative, and we would get
∫(1∕𝑥)𝑑𝑥 = ln |𝑥| + 𝐶

However, in this case we are integrating ln(𝑥)∕𝑥, which is undefined for 𝑥 ≤ 0, and we can assume that the primitive function is also undefined there (although technically it doesn't have to be undefined, only non-differentiable).
• Would you also be able to say that dx = x du for the first problem, and then use that to cancel out 1/x since you're multiplying by the opposite x/1 du, and you would be left with u^10 du? Or is that not the proper way of thinking about it?
(1 vote)

## Video transcript

- [Instructor] What we're going to do in this video is get some more practice identifying when to use u-substitution and picking an appropriate u. So let's say we have the indefinite integral of natural log of x to the, to the 10th power, all of that over x dx. Does u-substitution apply, and if so how would we make that substitution? Well the key for u-substitution is to see, do I have some function and its derivative? And you might immediately recognize that the derivative of natural log of x is equal to one over x. To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x dx. Now it becomes clear. We have some function, natural log of x, being raised to the 10th power, but we also have its derivative right over here, one over x. So we could make the substitution. We could say that u is equal to the natural log of x. And the reason why I picked natural log of x is because I see something, I see its exact derivative here, something close to its derivative, in this case its exact derivative. And so then I could say, du dx, du dx is equal to one over x, which means that du is equal to one over x dx. And so here you have it. This right over here is du, and then this right over here is our u. And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power du. And so you would evaluate what this is, find the antiderivative here, and then you would back-substitute the natural log of x for u, to actually evaluate this indefinite integral. Let's do another one. Let's say that we have the integral of, let's do something, let's do something interesting here. Let's say the integral of tangent x dx. Does u-substitution apply here? And at first you say well I just have a tangent of x, where is its derivative? But one interesting thing to do is well we can rewrite tangent in terms of sine and cosine. So we can write this as the integral of sine of x over cosine of x dx. And now you might say well where does u-substitution apply here? Well there's a couple of ways to think about it. You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. But more interesting you could say the derivative of cosine of x is negative sine of x. We don't have a negative sine of x, but we can do a little bit of engineering. We can multiply by negative one twice. So we could say the negative of the negative sine of x, and I stuck one of the you could say negative ones outside of the integral, which comes straight from our integration properties. This is equivalent. I can put a negative on the outside, a negative on the inside, so that this is the derivative of cosine of x. And so now this is interesting. In fact let me rewrite this. This is going to be equal to negative, the negative integral, of one over cosine of x times negative sine of x dx. Now does it jump out at you what our u might be? Well I have a cosine of x in the denominator, and I have its derivative, so what if I made u equal to cosine of x? u is equal to cosine of x, and then du dx would be equal to negative sine of x. Or I could say that du is equal to negative sine of x dx. And just like that I have my du here, and this of course is my u. And so my whole thing has now simplified to, it's equal to, the negative indefinite integral of one over u, one over u du. Which is a much easier integral to evaluate, and then once you evaluate this, you back-substitute cosine of x for u.