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Thread: find integral w/ trig

  1. #1
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    find integral w/ trig

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

    How should I go about this?
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    Re: find integral w/ trig

    Try $\displaystyle u = \sin x \implies \frac{du}{dx}= \dots$
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    Re: find integral w/ trig

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

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

    u = sinx
    it cant be this
    $\displaystyle \int \frac{u-u + cosx}{u^2} dx$
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    Re: find integral w/ trig

    ahh $\displaystyle cosx = \sqrt{ -u^2+1}$ via the trig identity, doh!
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    Re: find integral w/ trig

    fantastic, I got the solution now. thanks!
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  6. #6
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    Re: find integral w/ trig

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

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

    So the integral becomes

    $\displaystyle \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 \displaystyle \begin{align*} \cos{x} \end{align*} $ or any of its variants in the integrand...
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    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 \displaystyle \begin{align*} u = \sin{x} \end{align*} $ then $\displaystyle \displaystyle \begin{align*} du = \cos{x}\,dx \end{align*} $

    So the integral becomes

    $\displaystyle \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 \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.

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

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


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


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

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

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

    so now I have ....

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

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



    Using $\displaystyle 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

    $\displaystyle \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, $\displaystyle \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, $\displaystyle cosx= \sqrt{1-sin^2x} $ I now have

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

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

    $\displaystyle \int u^{-2} du$
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    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.

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

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


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


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

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

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

    so now I have ....

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

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



    Using $\displaystyle 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

    $\displaystyle \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, $\displaystyle \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, $\displaystyle cosx= \sqrt{1-sin^2x} $ I now have

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

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

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

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

    Can you see that $\displaystyle \displaystyle \begin{align*} \cos{x}\,dx \end{align*} $ appears in your integral? Just change it to $\displaystyle \displaystyle \begin{align*} du \end{align*} $...
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    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 \displaystyle \begin{align*} u = \sin{x} \end{align*} $ then $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = \cos{x} \implies du = \cos{x}\,dx \end{align*} $

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

    I agree. The more I do, the easier it will be for me to identify patterns.
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  10. #10
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    Re: find integral w/ trig

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

    $\displaystyle \int f'(x)[f(x)]^n}~dx=\frac{[f(x)]^{n+1}}{n+1}+C$? 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, $\displaystyle f(x)=\sin(x)$, $\displaystyle n=-2$ and $\displaystyle f'(x)=\cos(x)$

    Edit: You could also use the fact that $\displaystyle \frac{d}{dx}\csc(x)=\csc(x)\cot(x)$ and simply change the sign appropriately.
    Last edited by Quacky; Dec 6th 2011 at 05:40 PM.
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