Antiderivative of product of trig functions

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Homework Help Overview

The problem involves finding the antiderivative of the product of trigonometric functions, specifically sin^4(x)tan^2(x). The context is within a theoretical calculus course, with participants exploring integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods, including integration by parts and converting the integrand into sine and cosine terms. There are attempts to simplify the integral and questions about the effectiveness of different approaches.

Discussion Status

Several participants have offered suggestions and alternative methods for tackling the integral. There is an ongoing exploration of different strategies, with some expressing uncertainty about their previous attempts and seeking clarification on the approaches suggested by others.

Contextual Notes

Participants note the theoretical nature of their course and the timing of the discussion, as they are currently on break. There is mention of potential missing information regarding trigonometric identities that may be relevant to the problem.

snipez90
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Homework Statement


Find the antiderivative of [tex]sin^4(x)tan^2(x)[/tex]


Homework Equations


Trig identities I may have overlooked.

The Attempt at a Solution


I tried writing the integrand in terms of sin and cos but that didn't seem to lead anywhere. I tried integration by parts since the antiderivative of [tex]tan^2(x)[/tex] is [tex]tanx - x + C[/tex], but that approach lead to a more difficult integral that might have been solved using a messier integration by parts.

This isn't really homework since I'm in a theoretical calc course and we begin the theory of integration next quarter (in fact, we're on break). Anyways I used to be pretty good with elementary terms integration but I'm stuck. Could someone suggest a good first step? Thanks.
 
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I think this will go with integration by parts.
Try u = tan^2(x)sin^3(x) and dv = sin(x)dx.

This gave me an unwieldy-looking integral, but after simplifying to sines and cosines, I was left with [tex]\int \frac{sin x}{cos^2 x} dx[/tex] and [tex]\int sin^4 x dx[/tex]. There were constant multipliers that I have omitted.

The first integral is an easy one and the second isn't too hard.
Hope that helps.
 
Another potential way to do this integral is by converting everything to cosines because there is a cosine downstairs:

[tex]\int \sin^4x\,\tan^2x\,dx = \int\frac{(1-\cos^2 x)^3}{\cos^2 x}\,dx = \int \frac{1 - 3\cos^2 x + 3\cos^4 x - \cos^6 x}{\cos^2 x}\,dx[/tex]​

Each of the resulting integrals should be fairly simple to solve.
 
Thank you both for the suggestions.

Mark44, did you arrive at those two integrals after applying integration by parts once? I got sin sin^4(x) term but the the one before that (I think it was sin^4(x)/cos^2(x) or something). I'll try it again though.

Tedjn, I thought of the very first equality you posed but did not think to expand it. That is quite a neat trick, i.e., decrease the powers by expansion. A few expansions and some computation and it worked out well. Thanks.
 
snipez90 said:
Thank you both for the suggestions.

Mark44, did you arrive at those two integrals after applying integration by parts once?
Yes.
snipez90 said:
I got sin sin^4(x) term but the the one before that (I think it was sin^4(x)/cos^2(x) or something). I'll try it again though.

Tedjn, I thought of the very first equality you posed but did not think to expand it. That is quite a neat trick, i.e., decrease the powers by expansion. A few expansions and some computation and it worked out well. Thanks.
 

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