Results 1 to 10 of 10

Math Help - Integration - powers of trig functions

  1. #1
    Member
    Joined
    May 2008
    Posts
    138

    Integration - powers of trig functions

     \int 33\frac{cos^5(x)}{\sqrt{sin(x)}}

    Should I begin by splitting up the numerator? Multiplying by the conjugate? I've tried those two and some others to no avail.

    THanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by silencecloak View Post
     \int 33\frac{cos^5(x)}{\sqrt{sin(x)}}

    Should I begin by splitting up the numerator? Multiplying by the conjugate? I've tried those two and some others to no avail.

    THanks
    Put
     <br />
sin(x) = t <br />

    cos(x)dx = dt



     33\int \frac{cos^5(x)dx}{\sqrt{sin(x)}}


    =  33\int \frac{cos^4(x)dt}{\sqrt{t}}


    Now
     cos^4(x)

    =  (1 -sin^2(x))^2

    = (1 - t^2)^2

    = 1 + t^4  - 2t^2

    Hence integration becomes

    =  33\int \frac{(1 + t^4  - 2t^2 )dt}{\sqrt{t}}

    Now go ahead and ask incase of trouble
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    138
    Interesting...makes sense...I wish I could see this kind of stuff as I'm working the problems

    Edit: now thats not becoming a trig substitution is it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by silencecloak View Post
    Interesting...makes sense...I wish I could see this kind of stuff as I'm working the problems

    Edit: now thats not becoming a trig substitution is it?
    First of all
    Congrats on Becoming a Senior Member and on your 100 posts


    Yes its trignometric substitution
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2008
    Posts
    138
    Quote Originally Posted by ADARSH View Post
    First of all
    Congrats on Becoming a Senior Member and on your 100 posts


    Yes its trignometric substitution

    Thanks haha

    Is there another way to do this problem without trig sub? The chapter this homework is for has not covered that topic yet.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by silencecloak View Post
    Thanks haha

    Is there another way to do this problem without trig sub? The chapter this homework is for has not covered that topic yet.
    On a second thought, its something similar to your last question
    Last edited by ADARSH; February 24th 2009 at 01:43 AM. Reason: ,...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2008
    Posts
    138
    Quote Originally Posted by ADARSH View Post
    On a second thought, its something similar to your last question
    By last question, do you mean this post?

    http://www.mathhelpforum.com/math-he...ig-powers.html
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Yes If you do it that way , you should to do this one as well
    Reason: You did some similar substitution there
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2008
    Posts
    138
    Quote Originally Posted by ADARSH View Post
    Yes If you do it that way , you should to do this one as well
    Reason: You did some similar substitution there
    Ok, I will go back and look at this problem with that technique in mind, darn algebra
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    May 2008
    Posts
    138
    33\int  sin^{-1/2}(x)cos^4(x)cos(x)dx

    33\int  sin^{-1/2}(x)[1-sin^2(x)]^2cos(x)dx

    u = sin(x)
     du = cos(x)

    33\int u^{-1/2}(1-u^2)^2du


    Is this the right way?



    Edit: It's the right way, thanks for the help
    Last edited by silencecloak; February 24th 2009 at 02:57 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Integrals Involving Powers of Trig Functions.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 28th 2010, 12:26 AM
  2. Integration with trig functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 24th 2009, 04:08 AM
  3. trig powers integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 15th 2009, 02:46 PM
  4. Integration of trig functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 14th 2009, 10:45 PM
  5. integration with trig functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 6th 2006, 06:50 PM

Search Tags


/mathhelpforum @mathhelpforum