How to Convert Sine and Cosine to Secant for Integration?

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Homework Help Overview

The discussion revolves around techniques of integration, specifically focusing on the conversion of sine and cosine functions to secant in the context of an integration problem. The original poster expresses confusion regarding a particular transformation in the solution process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the double angle formula for sine and question the conversion of the product of sine and cosine into a secant function. There is also a note about the absence of an attachment that was initially referenced.

Discussion Status

Some participants are clarifying the mathematical concepts involved, while others are seeking to understand the specific transformation mentioned. There is an ongoing exploration of the definitions and relationships between the trigonometric functions involved.

Contextual Notes

One participant notes the lack of an attachment that was supposed to accompany the original post, which may limit the context for the discussion.

abot
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techniques of inegration, please help

in the attachment there is a problem related to techniques of integration, i fully understand everything until near the end of the problem they answer says

(1/6)o - (1/12)sin2o + C = (1/6)o - (1/6)sino coso + C

and then they convert the sinocoso to sec-1

this part is really confusing...can you help me please...

thank you
 
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abot said:
In the attachment there is a problem related to techniques of integration, i fully understand everything until near the end of the problem they answer says
(1/6)o - (1/12)sin2o + C = (1/6)o - (1/6)sino coso + C
?
But... there's no attachment? :confused:
 
Do you know the "double angle formula":
[tex]sin(2\theta)= 2sin(\theta)cos(\theta)[/tex]

That was what was used.

I'm not clear what you mean by "convert the sin([itex]\theta[/itex])cos([itex]\theta[/itex]) to sec-1"
Of course, sec([itex]\theta[/itex]) is defined as [itex]\frac{1}{cos(\theta)}[/itex] so if that "-1" is meant as reciprocal rather than "inverse function", it is true that
[tex]sin(\theta)cos(\theta)= \frac{sin(\theta)}{sec(\theta)}[/tex]
 

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