How do I use trig substitution to solve this integral?

In summary, the conversation discusses the method for solving the integral \int x(81-x^2)^{5/2}dx using trigonometric substitution. The final answer is given as -\frac{(\sqrt{81-x^2})^7}{7}+C, which can also be obtained using a table or inspection.
  • #1
p53ud0 dr34m5
94
0
im hoping i worked this out right; its long:
[tex]\int x(81-x^2)^{5/2}dx[/tex]
the integral contains [itex]a^2-x^2[/itex], so i set [itex]x=asin\theta[/itex]. that would make [itex]x=9sin\theta[/itex] and [itex]dx=9cos\theta d\theta[/itex]:
[tex]\int 9sin\theta(81-81sin^2\theta)^{5/2}9cos\theta d\theta = \int 9sin\theta[81(1-sin^2\theta)]^{5/2}9cos\theta d\theta[/tex]
the integral now contains [itex]1-sin^2\theta =cos^2\theta[/itex]
[tex]\int 9sin\theta(9cos\theta)^59cos\theta d\theta[/tex]
i used u-sub by setting [itex]u=9cos\theta[/itex] and [itex]du=-9sin\theta d\theta[/itex].
[tex]-\int u^6 du=-\frac{u^7}{7}+C[/tex]
i plugged my u back in:
[tex]-\frac{(9cos\theta)^7}{7}+C[/tex]
then, i drew my little triangle.
[tex]cos\theta=\frac{\sqrt{81-x^2}}{9}[/tex]
i plugged that into the [itex]cos\theta[/itex] and simplified and came out with:
[tex]-\frac{(\sqrt{81-x^2})^7}{7}+C[/tex]
that's my answer
 
Physics news on Phys.org
  • #2
You could also try the substitution u = 81 - x^2, which should be significantly easier.
 
  • #3
I punched it into mathematica and it produced the same as yours, save for a 7/2 exponent rather than 7.
 
  • #4
7/2 exponent is like my square root to the 7th
[tex]-\frac{(\sqrt{81-x^2})^2}{7}=-\frac{(81-x^2)^{7/2)}{7}[/tex]
also, we had to use trig substitution, so that left out the easy 81-x^2 sub
 
  • #5
p53ud0 dr34m5 said:
im hoping i worked this out right; its long:
[tex]\int x(81-x^2)^{5/2}dx[/tex]

It doesn't have to be.

You shouldn't have to use a trig or u substitution. Inspection isn't too hard with something like this.

[tex]\frac{(2)}{(-2)(7)}\int (-2x)(\frac{7}{2})(81-x^2)^{5/2}dx[/tex]
[tex]= -\frac{1}{7} (\sqrt {81-x^2})^7[/tex]
 
Last edited by a moderator:
  • #6
*Cough* Table *Cough*

Excuse me... something in my throat--- tables are great to have around so you barely have to think.
 
  • #7
Guess having someone else do it is the easiest way. :smile:
 

1. What is trig substitution?

Trig substitution is a technique used in calculus to simplify integrals involving trigonometric functions. It involves replacing a complicated trigonometric expression with a simpler one, often using a trigonometric identity.

2. When should I use trig substitution?

Trig substitution is typically used when trying to integrate functions involving sqrt(a^2 - x^2), sqrt(x^2 + a^2), and sqrt(x^2 - a^2). These types of integrals are usually solved more easily using trig substitution.

3. How do I know which trig substitution to use?

The substitution you use depends on the form of the integral. For sqrt(a^2 - x^2), you would use x = a sin(theta). For sqrt(x^2 + a^2), you would use x = a tan(theta). For sqrt(x^2 - a^2), you would use x = a sec(theta). It's important to also consider the limits of integration when choosing the appropriate substitution.

4. Are there any special cases where trig substitution wouldn't work?

Yes, there are certain cases where trig substitution may not work. For example, if the integral contains an odd power of sin or cos, or if the limits of integration are not compatible with the chosen substitution. In these cases, other integration techniques may be necessary.

5. Can I use trig substitution to solve indefinite integrals?

Yes, trig substitution can also be used to solve indefinite integrals. However, it's important to remember to include the appropriate constant of integration when finding the antiderivative of the trigonometric expression.

Similar threads

Replies
3
Views
935
  • Calculus
Replies
29
Views
515
Replies
2
Views
179
Replies
4
Views
189
  • Calculus
Replies
4
Views
804
  • Calculus
Replies
6
Views
1K
  • Calculus
Replies
1
Views
1K
  • Calculus
Replies
2
Views
2K
Back
Top