Trigonometric substitution

  • Thread starter tony873004
  • Start date
  • #1
tony873004
Science Advisor
Gold Member
1,752
143
This is an example from the book. Evaluate [tex]
\int {\frac{{\sqrt {9 - x^2 } }}{{x^2 }}dx}
[/tex]

I understand all the steps that get me up to [tex] = - \cos \theta \, - \theta \, + C[/tex]

Then the book goes on to explain:
"Since this is an indefinate integral, we must return to the original variable x. This can be done either by using trig identities to express cot theta in terms of sin theta=x/3 or by drawing a diagram, as in Fig. 1, where theta is intrepreted as an angle of a right triangle. Since sin theta = x/3....."

Where does this x/3 come from? The steps that follow make sense, but how did we know to just pull an x/3 out of the air and start using it?


Applying this to a homework question [tex]\int_{}^{} {x^3 \sqrt {9 - x^2 } } dx[/tex] I come up with [tex]243\left( {\frac{{\cos ^5 \theta }}{5} - \frac{{\cos ^3 \theta }}{3}} \right) + C[/tex]

Now, switching it back to x, I'm going to guess that [tex]{\sqrt {9 - x^2 } }[/tex] becomes the adjacent side of my triangle, but for no good reason other than the example of the book pulled the sqrt() out of the original equation and did the same.

I just don't get this triangle stuff. The lecture didn't do it for me, and the book's explanation doesn't do it for me either >:-(

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
VietDao29
Homework Helper
1,424
3
This is an example from the book. Evaluate [tex]
\int {\frac{{\sqrt {9 - x^2 } }}{{x^2 }}dx}
[/tex]

I understand all the steps that get me up to [tex] = - \cos \theta \, - \theta \, + C[/tex]

Then the book goes on to explain:
"Since this is an indefinate integral, we must return to the original variable x. This can be done either by using trig identities to express cot theta in terms of sin theta=x/3 or by drawing a diagram, as in Fig. 1, where theta is intrepreted as an angle of a right triangle. Since sin theta = x/3....."

Where does this x/3 come from? The steps that follow make sense, but how did we know to just pull an x/3 out of the air and start using it?

Nope, in the early steps of the solution, the author used the substitution:

[tex]x = 3 \sin \theta[/tex], so, rearrange it a bit, we'll have: [tex]\sin \theta = \frac{x}{3}[/tex]


Applying this to a homework question [tex]\int_{}^{} {x^3 \sqrt {9 - x^2 } } dx[/tex] I come up with [tex]243\left( {\frac{{\cos ^5 \theta }}{5} - \frac{{\cos ^3 \theta }}{3}} \right) + C[/tex]

Ok, so far, so good.

Now, switching it back to x, I'm going to guess that [tex]{\sqrt {9 - x^2 } }[/tex] becomes the adjacent side of my triangle, but for no good reason other than the example of the book pulled the sqrt() out of the original equation and did the same.

I just don't get this triangle stuff. The lecture didn't do it for me, and the book's explanation doesn't do it for me either >:-(

Well, no, you have:
[tex]3 \sin \theta = x, \quad \theta \in \left[ -\frac{\pi}{2} ; \ \frac{\pi}{2} \right][/tex]

[tex]\Rightarrow \sin \theta = \frac{x}{3}[/tex]

We'll now try to express cos(theta) in terms of x, since we have: [tex]\theta \in \left[ -\frac{\pi}{2} ; \ \frac{\pi}{2} \right][/tex], that means theta is in the first, and forth quadrant. So, it's cosine value will be positive.

Pythagorean Theorem states that: [tex]\cos ^ 2 \theta + \sin ^ 2 \theta = 1[/tex], so:

[tex]\Rightarrow \cos \theta = \sqrt{1 - \sin ^ 2 \theta} = \sqrt{1 - \frac{x ^ 2}{9}}[/tex].

Now, sub this to the expression above, and it's done.

Can you get it? Is it clear? :)
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
965
You keep avoiding mentioning the original substitution: [itex]3 sin(\theta)= x[/itex] or [itex]sin(\theta)= x/3[/itex]!
You solution involves [itex]cos(\theta)[/itex]. In order to convert back to the original x, you need to figure out what [itex]cos(\theta)[/itex] is from knowing what [itex]sin(\theta)[/itex] is. Since sine is defined as "opposite side divided by hypotenuse", imagine (or better draw) a right triangle with "opposite side" of length x and "hypotenuse" of length three. You can use the Pythagorean theorem to find that the length of the "near side" is [itex]\sqrt{9- x^2}[/itex] and, since cosine is "near side divided by hypotenuse", [itex]cos(\theta)= \sqrt{9-x^2}/3[/itex] which could also be written [itex]\sqrt{1- x^2/9}[/itex].
 
  • #4
tony873004
Science Advisor
Gold Member
1,752
143
Thanks for the replies. It makes a lot more sense now :)
 

Related Threads on Trigonometric substitution

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
775
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
0
Views
875
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
17
Views
2K
Top