Can you do integral of sec^5(x)tan^2(x) without reduction formula?

Click For Summary

Homework Help Overview

The problem involves finding the integral of sec^5(x)tan^2(x) without using a reduction formula. The subject area pertains to integral calculus, specifically focusing on trigonometric integrals.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers whether integration by parts could be a viable method and mentions attempts to split the integral in various ways. Another participant suggests rewriting the integral in terms of sec(x) and proposes specific choices for u and dv in the integration by parts process.

Discussion Status

Participants are actively exploring different approaches to the integral, with one providing a detailed suggestion on how to proceed using integration by parts. There is a sense of progress as one participant expresses appreciation for the insight shared.

Contextual Notes

There is a mention of the original poster's uncertainty about the method and the need to solve for the integral algebraically, indicating potential constraints in their understanding or approach.

aselin0331
Messages
7
Reaction score
0

Homework Statement



Can you do integral of sec^5(x)tan^2(x) without reduction formula?


Homework Equations





The Attempt at a Solution



If so, would it be integration by parts? I tried splitting it up in way too many ways to post them on here.

Thanks for any hints!
 
Physics news on Phys.org
aselin0331 said:

Homework Statement



Can you do integral of sec^5(x)tan^2(x) without reduction formula?


Homework Equations





The Attempt at a Solution



If so, would it be integration by parts? I tried splitting it up in way too many ways to post them on here.

Thanks for any hints!
First, get everything in terms of sec(x).
\int sec^5(x)tan^2(x)dx = \int sec^7(x) - sec^5(x)dx = \int sec^7(x)dx - \int sec^5(x)dx

Use integration by parts on both integrals on the right. For the first u = sec5(x) and dv = sec2(x) dx. For the second integral, u = sec3(x) and dv = sec2(x) dx.

I haven't taken this all the way through, but I'm reasonably sure it will work. You will probably need to solve for your integral algebraically.
 
You're fantastic! I was staring at Sec^7 and sec^5 for a while and that just clicked in!

Thanks
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K