Find the integral of sinx/cos^3x dx

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Homework Statement



Find the integral of sinx/cos^3x dx

Homework Equations





The Attempt at a Solution



How would I approach such a problem?
 
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Answers and Replies

  • #2
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If that's a cos(x) raised to the 3rd power in the denominator, then your integrand is just tan(x)sec^2(x) which is easy since the derivative of tan(x) is sec^2(x).
 
  • #3
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Wait so if I substitute U for cos(x)^3, then du=tan(x)sec^2(x)?
 
  • #4
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No, he's saying that the integrand can be rewritten as tanxsec^2 x which can be easily integrated.
 
  • #5
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the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??
 
  • #6
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the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??
The integrand of an integral is the expression between the summa [tex]\int[/tex] and the differential dx. The equation you wrote above is correct for the integrand, which you can now integrate easily.
 
  • #7
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the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??
Yes. If

[tex]y=\sec^2(x)[/tex]


Then:


[tex]\frac{dy}{dx} = \cdots [/tex]
 
  • #8
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tan(x)sec(x)?? But i don't understand, what happens to the sin(x) in the numerator..It seems like were just talking about the denominator.
 
  • #9
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tan(x)sec(x)?? But i don't understand, what happens to the sin(x) in the numerator..It seems like were just talking about the denominator.
Do you remember the common definition of tan(x) = sin(x)/cos(x)?
 
  • #10
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[tex] y=sec^2(x) \Rightarrow y=\frac{1}{cos(x)}\frac{1}{cos(x)}[/tex]

Use the product rule to differentiate that, you will see you have your derivative that is the [almost] the same as the integrand. Hence you have the answer.
 
  • #11
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wow, it finally makes sense! Thank you everyone!
 
  • #12
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Here is much simpler approach:

[tex]\int{\frac{sinx}{cos^3x} dx}[/tex]

u=cos(x)

du=-sin(x)dx

dx=-du/sin(x)

[tex]\int{\frac{sin(x)}{u^3}*\frac{-du}{sin(x)}}=[/tex]

[tex]=-\int \frac{du}{u^3}[/tex]

:wink:
 
  • #13
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This was more of the approach we learned in class. Would you then use the LN function?
 
  • #14
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You use [itex]\ln[/itex] if the integrand were 1/u where the denominator has a power of one, but for any other power, use the power rule for integrals.
 

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