#Can you try to solve this problem using trigonometric substitution?
109 messages · Page 1 of 1 (latest)
Factor out x^6. That should make it clear.
i did that
still came up with a diff answer from the textbook
can you try solving it for me using trig substitution
sqrt(x^6 - x^8)
= sqrt(x^6 * (1 - x^2))
= x^3 * sqrt(1 - x^2)
x = cos(t)
x^3 * sqrt(1 - x^2) = cos^3(t) * sin(t)```
I'm sure you can take it from there.
You used x = sin(t), which isn't wrong, but you forgot the substitution was inside a square root.
no
Since dx = cos(x)dx
so I have
x^3 sqrt (1-x^2) * dx
so it becomes sin^3 (x) * cos(x) * cos(x)
no?
Oh, right. I forgot to replace dx. In my defense, you didn't write it.
haha lol
When I fully expand it I end up with a weird answer very close to the textbook
but I cant tell where I deviated
from the correct answer
Okay, so we have int sin^3(t)cos^2(t) dt.
Yes
I then broke down sin^3(t) into sin^2(t)sin(t)
I converted sin^2(t) into (1-cos^2(t))
so I end up with
int (1-cos^2(t))*sin(t)*cos^2(t) dt
cos(t) = u
-sin(t) dt = du
dt = du/-sin(t)
-1 int (1-u^2)u^2 du
taking the antiderivative I get
-cos^3(t)/3 + cos^5(t)/5 + C
Gotta roll it back to terms of x.
plugging in the value of cos(t) through a triangle I end up with
-(1-x^2)^3/2 (1-x^2)^5/2 + C
-------------- + --------------
3x^3/2 5x^5/2
the textbook answer is
(-1+x^2)(2+3x^2)(sqrt x^6 - x^8)
---------------------------------------- + C
15x^3
Okay, wait, I can't read any of that. What's cos(t)?
(1-x^2)
...not according to my math.
Okay, wait, where did the xs in the denominator come from?
when I plug in the values in the right triangle, Cos (t) = adj/hyp so it is sqrt(1-x^2) / x
Why is your hypotenuse x?
because x = sin(t)
And sin(t) equals...?
You didn't even do the wrong math right. If your hypotenuse was x, then your adjacent side should've been sqrt(x^2 - 1).
ok
yea basically
I drew the triangle right and labeled
just wrote down the wrong thing
no the adj side should be sqrt (1-x^2)
...if x was the hypotenuse.
Hence "you didn't even do the wrong math right".
so my final answer is now
-(1-x^2)^3/2 (1-x^2)^5/2 + C
-------------- + --------------
3 5
Where the hell does the textbook get the xs in the denominator?
I am almost entirely certain the textbook is just wrong. I could be completely certain if I felt like differentiating that ridiculous monster.
Maybe its the same answer
just differently written
but this textbooks been giving me so many problems
It's not the same answer.
I've done so many questions
and came up with diff answers
and im certain that all the steps are took are legal
thats why I wanted to talk to someone on a call so we can try to figure it out and think of a new way of solving
,w d/dx (x^2 - 1)(2 + 3x^2)(sqrt(x^6 - x^8))/(15x^3)
Huh. Sonofabitch.
Hahaha
Perhaps they used U substitution however the book requires us to use trig substitution
,w d/dx ((1-x^2)^3/2)/3 + ((1-x^2)^5/2)/5
,w d/dx (1 - x^2)^(5/2)/5 - (1 - x^2)^(3/2)/3
it input the question wrong whatever
Which is also correct.
Or off by a constant, at least.
Hmm
Obvious first step is to combine the fractions. As in, so obvious I don't know why I'm the one who has to say it.
...multiply by x^3/x^3.
You did.
Hes the one who made my brain
im just messing around ik what you mean
thank you though
No prob.
I really have no clue what the hell's going on with the form of that answer.
All the questions and answers in the book is like that. The textbook has videos that answers each problem but the professor refuses to give us access to it and i tried finding them online
I emailed her and fought about it but she doesnt care
,w int sqrt(x^6 - x^8) dx
Okay, what the hell, WA?
wow
Okay, WA did u sub.
thats cheating