Results 1 to 10 of 10

Math Help - find integral w/ trig

  1. #1
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    find integral w/ trig

    I need help finding out how to Find \int \frac{cosx}{sin^2x} dx.

    How should I go about this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28

    Re: find integral w/ trig

    Try u = \sin x \implies \frac{du}{dx}= \dots
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Re: find integral w/ trig

    \int \frac{cosx}{u^2} dx

    How would I change the variable of the the numator so it is in term of u

    u = sinx
    it cant be this
    \int \frac{u-u + cosx}{u^2} dx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Re: find integral w/ trig

    ahh cosx = \sqrt{ -u^2+1} via the trig identity, doh!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Re: find integral w/ trig

    fantastic, I got the solution now. thanks!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,521
    Thanks
    1405

    Re: find integral w/ trig

    I seriously doubt you have the correct solution, as your solution process is completely off...

    If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} du = \cos{x}\,dx \end{align*}

    So the integral becomes

    \displaystyle \begin{align*} \int{\frac{\cos{x}}{\sin^2{x}}\,dx} &= \int{\frac{1}{u^2}\,du} \\ &= \int{u^{-2}\,du} \end{align*}

    Not anything which has a \displaystyle \begin{align*} \cos{x} \end{align*} or any of its variants in the integrand...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Re: find integral w/ trig

    Quote Originally Posted by Prove It View Post
    I seriously doubt you have the correct solution, as your solution process is completely off...

    If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} du = \cos{x}\,dx \end{align*}

    So the integral becomes

    \displaystyle \begin{align*} \int{\frac{\cos{x}}{\sin^2{x}}\,dx} &= \int{\frac{1}{u^2}\,du} \\ &= \int{u^{-2}\,du} \end{align*}

    Not anything which has a \displaystyle \begin{align*} \cos{x} \end{align*} or any of its variants in the integrand...
    Hello Prove it,

    i ended up with the same thing as you. Here is my reasoning.

    \int \frac{cosx}{sin^2x} dx

    u = sin(x) u'=cos(x)


    I must change the function in terms of u, so I start with the denominator.


    \int \frac{cosx}{u^2} dx

    Now I will change my numerator cosx, in terms of u. Using the trigonometric Identity sin^2x+cos^2x=1

    cosx= \sqrt{1-sin^2x} \Rightarrow  \sqrt{1-u^2}

    so now I have ....

    \int \frac{\sqrt{1-u^2}}{u^2} dx

    Now I must change the constant multiplier of dx. I am going to use \sqrt{1-u^2} dx



    Using dx= \sqrt{1-u^2} & du or u'=cosx, I set them equal to eachother. NOTE: I must also change u' in terms of u

    \sqrt{1-u^2} dx =cosx du   \Rightarrow dx =\frac{cosx du}{\sqrt{1-u^2} }  \Rightarrow \frac{dx}{du} =\frac{cosx }{\sqrt{1-u^2} }

    So now I have dx in terms of u, \frac{cosx }{\sqrt{1-u^2} } Notice that cosx still has an x. I must remember to keep everything in terms of u. So remembering back to the trigonometric identity, cosx= \sqrt{1-sin^2x} I now have

    \frac{\sqrt{1-u^2} }{\sqrt{1-u^2} } = 1
    I plug that back into my new integral

    \int \frac{1}{u^2} du

    \int u^{-2} du
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,521
    Thanks
    1405

    Re: find integral w/ trig

    Quote Originally Posted by delgeezee View Post
    Hello Prove it,

    i ended up with the same thing as you. Here is my reasoning.

    \int \frac{cosx}{sin^2x} dx

    u = sin(x) u'=cos(x)


    I must change the function in terms of u, so I start with the denominator.


    \int \frac{cosx}{u^2} dx

    Now I will change my numerator cosx, in terms of u. Using the trigonometric Identity sin^2x+cos^2x=1

    cosx= \sqrt{1-sin^2x} \Rightarrow  \sqrt{1-u^2}

    so now I have ....

    \int \frac{\sqrt{1-u^2}}{u^2} dx

    Now I must change the constant multiplier of dx. I am going to use \sqrt{1-u^2} dx



    Using dx= \sqrt{1-u^2} & du or u'=cosx, I set them equal to eachother. NOTE: I must also change u' in terms of u

    \sqrt{1-u^2} dx =cosx du   \Rightarrow dx =\frac{cosx du}{\sqrt{1-u^2} }  \Rightarrow \frac{dx}{du} =\frac{cosx }{\sqrt{1-u^2} }

    So now I have dx in terms of u, \frac{cosx }{\sqrt{1-u^2} } Notice that cosx still has an x. I must remember to keep everything in terms of u. So remembering back to the trigonometric identity, cosx= \sqrt{1-sin^2x} I now have

    \frac{\sqrt{1-u^2} }{\sqrt{1-u^2} } = 1
    I plug that back into my new integral

    \int \frac{1}{u^2} du

    \int u^{-2} du
    Your reasoning, though possibly correct, is extremely convoluted and difficult to follow.

    If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} \frac{du}{dx} = \cos{x} \implies du = \cos{x}\,dx \end{align*}

    Can you see that \displaystyle \begin{align*} \cos{x}\,dx \end{align*} appears in your integral? Just change it to \displaystyle \begin{align*} du \end{align*} ...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Re: find integral w/ trig

    Quote Originally Posted by Prove It View Post
    Your reasoning, though possibly correct, is extremely convoluted and difficult to follow.

    If you let \displaystyle \begin{align*} u = \sin{x} \end{align*} then \displaystyle \begin{align*} \frac{du}{dx} = \cos{x} \implies du = \cos{x}\,dx \end{align*}

    Can you see that \displaystyle \begin{align*} \cos{x}\,dx \end{align*} appears in your integral? Just change it to \displaystyle \begin{align*} du \end{align*} ...

    I agree. The more I do, the easier it will be for me to identify patterns.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    Re: find integral w/ trig

    Can I just point out a shorthand approach, using the formula:

    \int f'(x)[f(x)]^n}~dx=\frac{[f(x)]^{n+1}}{n+1}+C? Obviously, working at developing your ability to use the u-substitution is highly beneficial, so I'm not discouraging you from using it, but this approach would undoubtedly save you time and energy. Here, f(x)=\sin(x), n=-2 and f'(x)=\cos(x)

    Edit: You could also use the fact that \frac{d}{dx}\csc(x)=\csc(x)\cot(x) and simply change the sign appropriately.
    Last edited by Quacky; December 6th 2011 at 05:40 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 18th 2011, 01:12 AM
  2. Find the value of 'trig of inverse trig' questions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 9th 2010, 05:37 PM
  3. Replies: 1
    Last Post: February 17th 2010, 03:58 PM
  4. New to Trig. How to find cos tan and sin?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 30th 2009, 03:21 AM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

/mathhelpforum @mathhelpforum