Question

How can I use substitution to evaluate the integral from 0 to 3pi/8 of sin(2x-pi/4)dx?

Answer 1

You don't need substitution :)
Just a simple trick to remember. I'll explain...
But if you insist on sub, we can do that also.

Answer 2

The problem specifies substitution, but I would appreciate a short explanation on how to do it without as well, if you could :)

Answer 3

\[\large\rm \frac{d}{dx}[-\cos(2x)]\]Do you understand how to do this derivative? :)

Answer 4

Yes, with the chain rule, right? If I'm correct it would be -2sin2x.

Answer 5

-cosine goes back to sine
so yes, that's close :)

Answer 6

Yes, I realized my mistake right away

Answer 7

So if the derivative of -cos(2x) is 2sin(2x),
then we expect that the integral of sin(2x) will be `something like` cos(2x) right?
Maybe a negative, we're not sure exactly what is happening with the 2 as well.
But that is a good "guess" at this point, right?

Answer 8

Right

Answer 9

Anyway, I'll cut to the point.
Whenever you have these `linear coefficients`,
differentiation gives you an extra coefficient on the outside due to chain rule.

Think of integration doing the opposite.
You simply end up dividing by that linear coefficient.

Example:\[\large\rm \int\limits e^{2x}dx\quad=\quad \frac{1}{2}e^{2x}+C\]Differentiation would give us `multiplication of` 2 on the outside.
Integration gives us `division by` 2 on the outside.

Answer 10

Another example:\[\large\rm \int\limits \cos(7x+\pi)dx\quad=\quad \frac{1}{7}\sin(7x+\pi)\]Differentiation would normally give us an extra 7 on the outside,
so integration must be taking away a 7.

Answer 11

Ok, so that would mean the integral of sin(2x) would be -(cos(2x))/2, right?

Answer 12

Ok great!
We should probably do substitution to verify this process though, right? :)
And since the instructions told us to. hehe

Answer 13

Sorry if it is a little bit difficult to read, I'm not sure how to type it so that it will display it more clearly as an equation

Answer 14

Yea, substitution would be great ha

Answer 15

\[\large\rm \int\limits_0^{3\pi/8} \sin\left(\color{orangered}{2x-\frac{\pi}{4}}\right)\color{royalblue}{dx}\]

Answer 16

\[\large\rm \color{orangered}{u=2x-\frac{\pi}{4}}\]We want to take the derivative of our substitution,
and try to end up with the blue piece in our integral.

Answer 17

So du would be x^2, right?

Answer 18

Woops, differentiation :))

Answer 19

x's get smaller when differentiating, ya?

Answer 20

My bad, du = 2

Answer 21

\[\large\rm du=2dx\]Ok that's almost exactly what we wanted.
Maybe we can `solve` for dx here?

Answer 22

So dx would be du/2?

Answer 23

\[\large\rm \color{royalblue}{\frac{1}{2}du=dx}\]Ok great!

Answer 24

\[\large\rm \int\limits\limits_0^{3\pi/8} \sin\left(\color{orangered}{2x-\frac{\pi}{4}}\right)\color{royalblue}{dx}\quad=\quad \int\limits\limits \sin\left(\color{orangered}{u}\right)\color{royalblue}{\frac{1}{2}du}\]

Answer 25

Do you understand why I erased the limits?

Answer 26

I think so. The way it is written without the limits now, I can take the indefinite integral first, then evaluate it later, right?

Answer 27

Well the only reason I pointed that out is because
those old limits `were values for x`.
You can't plug them in for the variable u.

So yes, you have a couple options:
1. Find new limits for u, integrate, evaluate using new limits.
2. Integrate, undo substitution, use original limits.

Answer 28

I think I like using that second method better

Answer 29

I hope my example didn't mess you up from earlier.
We agreed that the antiderivative of sin(2x) was -(cos2x)/2.

But let's try to match it with this problem a little better.
sin(2x+stuff) we expect to give us -(cos(2x+stuff))/2

The inner function won't change,
just a quick note there :)

Answer 30

I think I'm following along alright so far., yes.

Answer 31

I can factor out the 1/2 and put it in front of the integral, right?

Answer 32

\[\large\rm =\frac12\int\limits \sin u~du\]Sounds good.

Answer 33

Thanks for the help :)

Answer 34

Got it from there? c:

Answer 35

Yes, it makes sense from here, thanks

Answer 36

cool