Fundamental Integral Concept

[BigNate]

Hello Everyone,

I just want to make sure I understand. Can someone please confirm the below statement (my understanding) is correct?

My understanding: in order to evaluate an integral, the integral needs to be in u*du form. If it is not exactly in u*du form, it NEEDS to be manipulated by mulitplying by something inside the integral and balancing this with the reciprocal outside the integral sign. All integrals need to be in u*du form before integrating always. Derivatives do not follow this rule.

Example: if I need to evaluate the integral of sin(2x) dx, I would set u=2x and du=2x. Because we only have a dx and are missing the 2, we would want to multiply the inside by 2 and put a 1/2 on the outside to balance this.

Is the above true? Any flaws to my understanding and logic?

Thanks!

 

[ksdhart2]

Well, what you've detailed is a good summary of the concept of u-substitution, except for one thing - it's not the reciprocal you multiply by, but rather the derivative of u. In your example, you've made what I suspect is a typo, but I'll use that as a starting point to hopefully explain it a bit clearer. You started with:

\(\displaystyle \displaystyle \int sin(2x) \: dx\)

Next, you used u-substitution. Let \(\displaystyle u=2x\). By the chain rule, that makes \(\displaystyle du=2 \: dx\) (Here's the part I suspect was a typo, you wrote 2x which is wrong). By using a bit of algebra, we can see that \(\displaystyle dx=\dfrac{1}{2} \: du\). Now let's make the appropriate substitutions:

\(\displaystyle \displaystyle \int sin(u) \cdot \dfrac{1}{2} \: du\)

We can then pull out the 1/2 out front, and solve the integral in terms of u. Now let's consider a different example to fully showcase that it is the derivative, not the reciprocal:

\(\displaystyle \displaystyle \int \sqrt{5x+7} \: dx\)

Let \(\displaystyle u=5x+7\), so \(\displaystyle du=5 \: dx\) and \(\displaystyle dx=\dfrac{1}{5} \: du\)

\(\displaystyle \displaystyle \int \sqrt{u} \cdot \dfrac{1}{5} \: du\)

Again, we can pull out the 1/5 and solve in terms of u.

That being said, u-substitution is just one of several ways of dealing with integrals. If your Calculus class is anything like mine was, you'll soon learn about integration by parts and other methods for dealing with more complicated trigonometric integrals. So, it's not true that integrals always have to be the form u du, although it can often help.

 

[BigNate]

Ahhhh, I see. Thank you for your time and your detailed explanation. Makes perfect sense!