Integration Using Trigonometric Substitution Help Needed!

  • #1
23
0

Homework Statement


Integral of $$ x^3\sqrt{x^2+16}dx $$
answer should give
$$ 1/5(x^2+16)^{5/2} -16/3(x^2+16)^{1/2}+C $$

Homework Equations


x=atanθ

The Attempt at a Solution



IMG_20150222_141431.jpg

Mod note: The integral is ##\int x^3 \sqrt{x^2 + 16} dx##
The published answer is ##1/5(x^2+16)^{5/2} -16/3(x^2+16)^{1/2}+C##

To the OP: Using $ $ (i.e., with a space between the $ symbols) causes the browser to not render what's between the $ pairs. For inline LaTeX, you can use # # (no space) at the beginning and end.
 

Attachments

  • IMG_20150222_141431.jpg
    IMG_20150222_141431.jpg
    34.3 KB · Views: 365
Last edited:

Answers and Replies

  • #3
23
0
I'm trying to find the integral of the equation, but it doesn't give the same answer as the one I'm supposed to get, and I don't know what I'm doing wrong. I'm using trigonometric substitution which lets x=4sinΘ.
 
  • #4
In the second step after you did the substitution, the whole things becomes,
4*((tan0^2)+1)^(1/2)=4sec0 and not 4sec0^2
 
  • #5
can you post the question? Like the integral you are trying to find the answer for?
 
  • #6
23
0
IMG_20150222_143359~2.jpg


No problem! I tried using Latex, but it doesn't work...It's #6
 

Attachments

  • IMG_20150222_141431.jpg
    IMG_20150222_141431.jpg
    52.4 KB · Views: 366
  • #7
I might be able to help you....I love integration...but I don't understand anything from the picture you posted
 
  • #8
35,237
7,057
can you post the question? Like the integral you are trying to find the answer for?
See my previous post. It's fairly clear which integral the OP is trying to find.
 
  • #10
23
0
Thank You Mark! I'm still learning to use Latex... I changed it and it should appear ok, now!
 
  • #12
23
0
Ok, I'm currently trying, correcting the mistake I made with my sec^2. It looks good!
 
  • #13
23
0
I can officially say that it works now! Thank you Faris! Such a "minor" mistake that really changed everything in the problem!
 
  • #14
I might be able to help you....I love integration...but I don't understand anything from the picture you posted

He meant Question 6 from the picture
 
  • #15
Your trig substitution is correct. When you square root, you should get 4sec(x) (I'll use x instead of theta for simplicity)
That will leave you with,
64tan3(x) * 4sec(x)

or simply
256tan3(x)*sec(x)

Break the tan3(x) up, you get
256tan2(x)*sec(x)*tan(x)

I wont do the rest for you but I'll give you a clue. It involves a trig identity and a u substitution
 
  • #16
23
0
Thank you! It works now, my mistake was simply that I forgot to take the sqrt of sec^2
 
  • #17
hey, if trigonometric substitution is not necessary, it is much easier to put x^2+16=t.......=>x^2=t-16.......dx=dt/2x.......x^3=x*x2.....
then integral is ##\int { x*(t-16)*\sqrt(t)*dt/2x} ## => ##\int {(t-16)\sqrt(t) dt/2} ## ............now expand normally...
 

Related Threads on Integration Using Trigonometric Substitution Help Needed!

Replies
8
Views
2K
Replies
34
Views
4K
  • Last Post
Replies
4
Views
2K
Replies
1
Views
585
Replies
2
Views
2K
Replies
9
Views
2K
Replies
4
Views
1K
Replies
3
Views
3K
Replies
5
Views
3K
  • Last Post
Replies
7
Views
914
Top