• Support PF! Buy your school textbooks, materials and every day products Here!

Integrating trig powers, why not by parts?

  • #1

Homework Statement



Please help me understand the reason for substituting various trig identities into trig functions with powers instead of integration by parts. Does integration by parts not work on trig functions with powers, or is it just so much work that substituting trig identities to reduce powers ends up making the evaluation much more efficient?

My current rationale is that some trig functions can be integrated by U-substitution if one term is a derivative of the other term within the integrand. So in these cases, substituting trig identities are not necessary.

Then there is the case where the integrand contains trig functions in which one is not the derivative of the other and thus U-substitution won't work. In this case, integration by parts should work since the point of integration by parts is to evaluate the integral of two unrelated functions, [(f(x))(g(x)], within the integral.

So then what is the reason for using trig identities to substitute into trig functions with powers? Will integration by parts not work in these cases or is it that integration by parts is just to cumbersome so subbing trig identities makes evaluating more manageable?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,224
947

Homework Statement



Please help me understand the reason for substituting various trig identities into trig functions with powers instead of integration by parts. Does integration by parts not work on trig functions with powers, or is it just so much work that substituting trig identities to reduce powers ends up making the evaluation much more efficient?

My current rationale is that some trig functions can be integrated by U-substitution if one term is a derivative of the other term within the integrand. So in these cases, substituting trig identities are not necessary.

Then there is the case where the integrand contains trig functions in which one is not the derivative of the other and thus U-substitution won't work. In this case, integration by parts should work since the point of integration by parts is to evaluate the integral of two unrelated functions, [(f(x))(g(x)], within the integral.

So then what is the reason for using trig identities to substitute into trig functions with powers? Will integration by parts not work in these cases or is it that integration by parts is just to cumbersome so subbing trig identities makes evaluating more manageable?
Wow!! This an awfully general question.
 
  • #3
6,054
390
Your question is too broad. You need to say what form of trig. expressions you have in mind. Otherwise that could be anything, including those that cannot be integrated in elementary functions.
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,705
1,722

Homework Statement



Please help me understand the reason for substituting various trig identities into trig functions with powers instead of integration by parts. Does integration by parts not work on trig functions with powers, or is it just so much work that substituting trig identities to reduce powers ends up making the evaluation much more efficient?

My current rationale is that some trig functions can be integrated by U-substitution if one term is a derivative of the other term within the integrand. So in these cases, substituting trig identities are not necessary.

Then there is the case where the integrand contains trig functions in which one is not the derivative of the other and thus U-substitution won't work. In this case, integration by parts should work since the point of integration by parts is to evaluate the integral of two unrelated functions, [(f(x))(g(x)], within the integral.

So then what is the reason for using trig identities to substitute into trig functions with powers? Will integration by parts not work in these cases or is it that integration by parts is just to cumbersome so subbing trig identities makes evaluating more manageable?


Homework Equations





The Attempt at a Solution

What do you mean by "trig functions with powers"? Is is something like ##\sin(x^p) \cos(x^q), ## or ##\sin^p(x) \cos^q(x)## or ## x^p \sin(x) + x^q \cos(x)?##

RGV
 
  • #5
33,075
4,779
What do you mean by "trig functions with powers"? Is is something like ##\sin(x^p) \cos(x^q), ## or ##\sin^p(x) \cos^q(x)## or ## x^p \sin(x) + x^q \cos(x)?##
I'm going to guess that he means expressions like ##sin^2(x)##, ##tan^3(x)## and the like.
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,508
730
I think a partial an answer to the OP is to remember that integration by parts comes by trying to invert the product rule. This is much like the situation where the u substitution comes from trying to reverse the chain rule. Depending on which form your integral is, one or the other of these methods might be more appropriate. But not all integrals are of either form and may not be amenable to either method.
 

Related Threads for: Integrating trig powers, why not by parts?

  • Last Post
Replies
11
Views
2K
Replies
8
Views
4K
Replies
1
Views
1K
Replies
12
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
3
Views
2K
Replies
6
Views
544
Top