Integrating trig powers, why not by parts?

In summary, the reason for using trig identities to substitute into trig functions with powers is to simplify the integrand and make it easier to evaluate using either integration by parts or u-substitution methods.
  • #1
LearninDaMath
295
0

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

 
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  • #2
LearninDaMath said:

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
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
LearninDaMath said:

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
Ray Vickson said:
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
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 to Integrating trig powers, why not by parts?

1. What is the purpose of integrating trig powers using the integration by parts method?

The integration by parts method is used to break down a complex integral into simpler parts, making it easier to solve. It can be particularly useful when dealing with trigonometric functions that are raised to a power.

2. Is there a specific reason why integration by parts is more efficient for trig powers?

Yes, integration by parts is more efficient for trig powers because it helps to reduce the power of the trigonometric function, making it easier to integrate. This is especially helpful when the power is an odd number.

3. Can you provide an example of integrating a trig power using the integration by parts method?

Sure, an example would be integrating ∫sin^2(x) dx. Using integration by parts, we can rewrite this as ∫sin(x)*sin(x) dx. Then, we can choose u=sin(x) and dv=sin(x) dx, and use the formula ∫u*dv = uv - ∫v*du. This will simplify the integral to ∫sin(x)*cos(x) dx, which can be easily solved using substitution.

4. Are there any specific rules or guidelines to follow when using integration by parts for trig powers?

Yes, when using integration by parts for trig powers, it is important to choose the trig function with the highest power as u, and the other trig function as dv. Additionally, if the power of the trig function is odd, it is important to use the trig identity sin^2(x)+cos^2(x)=1 to simplify the integral.

5. Can integration by parts be used for all types of trigonometric functions?

Yes, integration by parts can be used for all types of trigonometric functions, including sine, cosine, tangent, secant, cosecant, and cotangent. However, the approach may vary depending on the type of trigonometric function and its power.

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