Question

Solve the definite integral using u-substitution @zepdrix

Answer 1

Would I use u = sin x?

Answer 2

@zepdrix @tkhunny

Answer 3

No.
You would like to get rid of the addition under the root,
that would make things easier for you.

Answer 4

So you want more than just sinx for your u

Answer 5

so u = 4 + 3sin x?

Answer 6

good good good.

Answer 7

please try this substitution:
\(4+3 \sin x =u\)

Answer 8

k, so du/dx = 3cos x

Answer 9

yes!

Answer 10

then dx = \(\large\frac{1}{3\cos x}du\) yeah?

Answer 11

woah :o

Answer 12

is that a yes?

Answer 13

I'm not sure why you put the cosine on the side with the du,
the cosine involves x, yes?

Answer 14

huh?

Answer 15

Hmm you have a weird approach to your differential :))
Let's see..

Answer 16

so you can plug in for dx

Answer 17

\[\large\rm \frac{du}{dx}=3\cos x\]From here, sort of "multiply" dx to the other side,\[\large\rm du=3\cos x~dx\]Then divide by 3,\[\large\rm \frac{1}{3}du=\cos x~dx\]

Answer 18

after that substitution, we get:
\[\Large \int {\frac{{du/3}}{{\sqrt u }}} \]

Answer 19

Ah, I see. I never thought to do it that way

Answer 20

in response to zepdrix

Answer 21

so, you get \(\Large3\sqrt{4+3\sin x}\) right? before plugging in pi and -pi?

Answer 22

no, wait. A 6 in front, yeah? not a 3?

Answer 23

\[\large\rm \int\limits\frac{\color{royalblue}{\cos x~dx}}{\sqrt{\color{orangered}{4+3\sin x}}}\]And we made our substitution,\[\large\rm \color{orangered}{u=4+3\sin x}\]Which led to,\[\large\rm \color{royalblue}{\frac{1}{3}du=\cos x~dx}\]Giving us an integral of\[\large\rm \int\limits\limits\frac{\color{royalblue}{\frac{1}{3}du}}{\sqrt{\color{orangered}{u}}}\]Follow the colors ok?

Answer 24

no, I think I made a mitake with the fractions...

Answer 25

ok, following the colors

Answer 26

How do you do the colors?

Answer 27

Haha umm
well you can write some code in the equation tool :D
so for color you do \color{}{}

In the first curly brace you put your color,
I used orangered and royalblue for my colors, i like those.
and in the second braces you put your math.

Answer 28

\(\Large\int\color{royalblue}{\frac{1}{3}}\color{orangered}{u^{-1/2}}\color{royalblue}{du}\)

How's this?

Answer 29

lol nice XD
Ok so simple power rule from there, ya?

Answer 30

yeah, so \(\Large \color{royalblue}{\frac{2}{3}}\color{orangered}{\sqrt{u}}\)?

Answer 31

We don't really need the color anymore :) lol
that was just to help match up our `substitution pieces`.
But ok good.

From here, you can undo your substitution,
and use the original boundary values,
OR
find new boundary values.

Just make sure you don't plug pi and -pi into this u.

Answer 32

What do you mean find new boundary values?

Answer 33

Your integral had boundary/limit values of x=-pi and x=pi.
These do not get plugged into u.

Answer 34

I know. You plug in 4 + 3sin x for u, then pi and -pi for x

Answer 35

Undo substitution?
Ok good :) Seems like a good route to take.

Answer 36

Yes, but what's the other option? What do you mean by "OR find new boundary values. " ?

Answer 37

Does the definite integral = 0?

Answer 38

yay good job \c:/

Umm for "other option", it's not much different.
You just plug in your boundary values when you're doing your substitution.
So for this problem we have plugged in,\[\large\rm u=4+3\sin x\qquad\to\qquad u=4+3\sin(-\pi)\qquad\to\qquad u=4\]And for the other boundary value,\[\large\rm u=4+3\sin x\qquad\to\qquad u=4+3\sin(\pi)\qquad\to\qquad u=4\]Which would have given us new boundaries for our integral,\[\large\rm \int\limits_{u=4}^4 \frac{1}{3}u^{-1/2}du\]

Answer 39

Ah, I see. :)

Answer 40

You want to do that if plugging in the values at the very end creates a huge mess... it's just a way of breaking up the work I guess.

Answer 41

Thank you for the help again!

Answer 42

np