Question

Looking for insight onto solving this integral. I know it requires some trig sub and and completing the square but I cant fig it out! thanks

Answer 1

\[\int\limits_{}^{}\sqrt{8x-x^2}dx\]

Answer 2

Hi c:

Answer 3

Hey :D

Answer 4

Completing the square? Hmmm.
Personally I would start by taking the negative off of the square term.
I think it'll just be easier to work with.\[\Large\bf\sf \int\limits \sqrt{-(x^2-8x)}\;dx\]And then let's complete the square on stuff in the brackets.
That seem ok? :o

Answer 5

Yea I tried that I get something like sqrt(16-(x-4)^2)

Answer 6

Ah yes, very good.

Answer 7

We would like our thing under the root to resemble something like this:\[\Large\bf\sf 1-(stuff)^2\]Because then we can make a nice substitution,\[\Large\bf\sf stuff=\sin \theta\]And apply our square identity to simplify it down,\[\Large\bf\sf 1-(stuff)^2\quad=\quad 1-\sin^2\theta\quad=\quad \cos^2 \theta\]

Answer 8

But see how we have a 16 as our first value under the root?
Uh oh it's not a 1.
What can we do? :O

Answer 9

Factor out a four \[\int\limits_{}^{} and thats where im not sure how to proceed

Answer 10

We have to factor a 16 out of the square term as well,
I'm going to bring it into the square as a 4,
\[\Large\bf\sf 4\int\limits \sqrt{1-\left(\frac{x-2}{4}\right)^2}\;dx\]Does that make sense?

Answer 11

x-4.. woops lemme fix that real quick.

Answer 12

\[\Large\bf\sf 4\int\limits\limits \sqrt{1-\left(\frac{x-4}{4}\right)^2}\;dx\]

Answer 13

OKay and then would I let x=sin?

Answer 14

You let the big blob inside of the square = sin theta.

Answer 15

\[\Large\bf\sf \frac{x-4}{4}\quad=\quad \sin \theta\]

Answer 16

\[\Large\bf\sf 4\int\limits\limits\limits \sqrt{1-\left(\sin \theta\right)^2}\;dx\]Which cleans up the stuff under the root quite nicely, yes?

Answer 17

Yea that does seem to work, and than I assume you can write \[4\int\limits_{}^{}\cos \theta d \theta \]

Answer 18

=\[4\sin \theta \]

Answer 19

No, our differential doesn't simply switch from dx to dtheta.
He's going to give us some extra bagage with him.

Answer 20

oh i see

Answer 21

so i find \[d \theta \] by taking the deravtive of our x=4sin(theta)+4

Answer 22

dx=4cos(theta)d(theta)

Answer 23

Ok looks good.

Answer 24

\[16\int\limits_{}^{}(\cos \theta)^2d \theta \] ?

Answer 25

Ok good.

Answer 26

Remember your Cosine Half-Angle Identity?

Answer 27

yea i believe the integral comes to, (theta/2)+sin(2theata)/4

Answer 28

Ok good, with a 16 multiplying both terms yes?

Answer 29

Yea correct

Answer 30

So now we need our solution back in terms of x,
which will take a little bit of effort.

Answer 31

Theta doesn't have a nice x representation so for our first term we'll just attack it head on using our substitution from earlier.\[\Large\bf\sf \frac{x-4}{4}=\sin \theta\qquad\to\qquad \theta=\arcsin\frac{x-4}{4}\]

Answer 32

\[\Large\bf\sf 8\theta+4\sin2\theta+\mathcal C\]

So that will deal with the first term.
Understand how I came up with the inverse sine?

Answer 33

Yea because it is basically saying where does sin =

Answer 34

But I find it confusing how to go about making the triangle

Answer 35

Before we make a triangle,
we need our sin2theta to be in terms of the angle theta, not 2theta.

Remember Sine Double Angle? :o

Answer 36

Sin2theta=2sincos

Answer 37

Ok good,\[\Large\bf\sf 8\theta+8\sin\theta\cos\theta+\mathcal C\]

Answer 38

Looking back at our substitution,\[\Large\bf\sf \sin \theta\quad=\quad\frac{\color{royalblue}{x-4}}{4}\quad=\quad \frac{\color{royalblue}{opposite}}{hypotenuse}\]There is one way we can set up our triangle.

Answer 39

Understand how to label the sides?

Answer 40

Yup! but which sqrt do we use for the adjacent?

Answer 41

Use Pythagorean Theorem to find the adjacent side.

Answer 42

\[\Large\bf\sf b^2=4^2-(x-4)^2\]

Answer 43

Alright makes sense, and after that just plugging in would the give the final result?

Answer 44

Yes.
For this problem you'll need to plug in both for your sin(theta) and your cos(theta).

Shouldn't be too bad though.

Answer 45

Yup, thanks for the help

Answer 46

np

Answer 47

And for the theta alone, what do i use? arcsin?

Answer 48

@zepdrix

Answer 49

Yes, that's as good as we're going to get for our theta.\[\Large\bf\sf 8\arcsin\left(\frac{x-4}{4}\right)+8\sin\theta\cos\theta+\mathcal C\]

Answer 50

Yea so 8arcsin(x-4/4)+8(x-4/4)(8-x/4)?