1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the integral of sinx/cos^3x dx

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the integral of sinx/cos^3x dx

    2. Relevant equations



    3. The attempt at a solution

    How would I approach such a problem?
     
    Last edited by a moderator: Jul 9, 2014
  2. jcsd
  3. Sep 14, 2009 #2
    Re: Integration

    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).
     
  4. Sep 14, 2009 #3
    Re: Integration

    Wait so if I substitute U for cos(x)^3, then du=tan(x)sec^2(x)?
     
  5. Sep 14, 2009 #4
    Re: Integration

    No, he's saying that the integrand can be rewritten as tanxsec^2 x which can be easily integrated.
     
  6. Sep 14, 2009 #5
    Re: Integration

    the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??
     
  7. Sep 14, 2009 #6
    Re: Integration

    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.
     
  8. Sep 14, 2009 #7
    Re: Integration

    Yes. If

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


    Then:


    [tex]\frac{dy}{dx} = \cdots [/tex]
     
  9. Sep 14, 2009 #8
    Re: Integration

    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.
     
  10. Sep 14, 2009 #9
    Re: Integration

    Do you remember the common definition of tan(x) = sin(x)/cos(x)?
     
  11. Sep 14, 2009 #10
    Re: Integration

    [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.
     
  12. Sep 14, 2009 #11
    Re: Integration

    wow, it finally makes sense! Thank you everyone!
     
  13. Sep 15, 2009 #12
    Re: Integration

    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:
     
  14. Sep 15, 2009 #13
    Re: Integration

    This was more of the approach we learned in class. Would you then use the LN function?
     
  15. Sep 15, 2009 #14
    Re: Integration

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find the integral of sinx/cos^3x dx
  1. Integrate cos(lnx)dx (Replies: 12)

  2. Integral (x e^-3x dx) (Replies: 2)

  3. Finding ∫x sin^3x dx (Replies: 4)

Loading...