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# 𝘶-substitution: definite integrals

When using 𝘶-substitution in definite integrals, we must make sure we take care of the boundaries of integration.

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• There seem to be some questions that DON'T change the boundaries of integration. How do you tell when or when not to change the boundaries? • If you stick with the anti-derivative as a function of u, then you need to change the bounds of integration.
If you "back substitute" for u, then that's when you don't change the bounds.
The choice (back substitute or not) is up to you.

The closing section of the video covers this:
You can either keep it a definite integral [as a function of u]
and then change your bounds of integration
and express them in terms of u.
That's one way to do it.
The other way is to try to evaluate
the indefinite integral,
use u-substitution as an intermediary step,
then back-substitute back and then evaluate
at your bounds. [values of x]
• I'm having difficulties conceptualizing the difference between definite and indefinite integrals, in particular the idea of the constant, +C, used in the indefinite integrals. At around , Sal integrates the u-substitution in the usual fashion and it makes sense that he uses the boundaries x = 2 to x = 1 because the problem is a definite integral. I guess my question is if you integrated the u-substitution as an indefinite integral you would get (u^4)/4 + C but the C goes away when you've constricted it to a set of boundaries. If there was a constant in your original problem, say a +4 for example, wouldn't the constant have some sort of effect on your answer even after evaluating at the min and max boundaries? I'm not sure if I'm explaining the question effectively but I appreciate any clarification on the topic. • What happened to the 2x in the original integral? • What is the difference between definite and indefinite integrals? • A definite integral tells you the area under the curve between two points a & b, and indefinite integral gives you the general form of the anti-derivative of the function. Operationally the only difference is plugging in values once you've integrated. Sometimes it's convenient to write down the indefinite form, and then plug in values a, b so that you can keep track if you need to change your bounds.
• I don't understand why the integral of secant is tangent. I got a quiz asking this, which doesn't explain about the trigonometric integral. May I get some explanation? • The integral of secant(x) is ln(secant(x) + tangent(x)), not tangent(x). I assume you are asking why the integral of secant(x)^2 is tangent(x). Seeing why the integral of secant(x)^2 equals tangent(x) is easier if you ask why is the derivative of tangent(x) equal to secant(x)^2.

tan(x) = sin(x)/cos(x)
1. (tan(x))' = (sin(x)/cos(x))' Use the derivative quotient rule
2. (tan(x))' = (cos(x)^2+sin(x)^2)/cos(x)^2 Split the fraction into two terms and simplify
3. (tan(x))' = 1+sin(x)^2/cos(x)^2 Substitute tan(x) for sin(x)/cos(x)
4. (tan(x))' = 1+tan(x)^2 Substitute the expression using the identity tan(x)^2+1=sec(x)^2
5. (tan(x))' = sec(x)^2
• if A definite integral tells us the area under the curve between two points a & b can we say that an indefinite integral tells us the area under the curve between +infinity and -infinity? • What if substituting the bounds of integration makes the bottom one larger than the top? • Which method of u substitution is preferable in most circumstances? • Why is the integral of 1/(1-u^2)^1/2= sin^-1(u^2)+C? • At , why does substituting the "x" bounds of integration into the u equation help you find the new bounds in respect to u? I still don't understand the logic or reasoning behind this. • We let u=x^2+1.
Since we can calculate u given an x value, we can calculate what the u-bounds are given the x-bounds (since were are changing the variable of integration from dx to du, we need to change the bounds from x-bounds to u-bounds).
First, plug the lower bound x=1 into the equation. You get u=(1)^2+1, so the lower bound of our rewritten integral is u=2.
Plug the upper bound x=2 into the same equation and get u=2^2+1, so the upper bound for the rewritten integral is u=5.