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Math Help - can someone please help with this trig subsitution?

  1. #1
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    can someone please help with this trig subsitution?

    can someone please help with this trig subsitution?-trig-sub.jpg


    Thanks!
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  2. #2
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    Just to clarify: you're asked to compute

    \displaystyle{\int\frac{du}{u\sqrt{5-u^{2}}}}.

    Is that correct?
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  3. #3
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    Quote Originally Posted by softballchick View Post
    Click image for larger version. 

Name:	trig sub.jpg 
Views:	21 
Size:	10.0 KB 
ID:	18447


    Thanks!
    It would make things easier if you typed your questions out rather than using an attachment. There is nothing in your attachment that cannot be easily typed. Also, it would be easier to provide specific help if you showed what you had tried and said where you are stuck.
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  4. #4
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    Quote Originally Posted by softballchick View Post
    Click image for larger version. 

Name:	trig sub.jpg 
Views:	21 
Size:	10.0 KB 
ID:	18447

    \int_{\frac{1}{u\sqrt{5-u^2}}du


    Thanks!
    One way is to draw a right-angled triangle, with perpendicular sides u,\ \sqrt{5-u^2}

    and hypotenuse \sqrt{5}

    then

    cos\theta=\frac{u}{\sqrt{5}}\ \Rightarrow\ u=\sqrt{5}cos\theta

    du=-\sqrt{5}sin\theta\ d\theta

    sin\theta=\frac{\sqrt{5-u^2}}{\sqrt{5}}\ \Rightarrow\ \sqrt{5-u^2}=\sqrt{5}sin\theta

    then use these to write a trigonometric integral
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  5. #5
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    \int{\frac{du}{u\sqrt{5 - u^2}}}.

    Let u = \sqrt{5}\sin{\theta} so that du = \sqrt{5}\cos{\theta}\,d\theta and the integral becomes

    \int{\frac{\sqrt{5}\cos{\theta}\,d\theta}{\sqrt{5}  \sin{\theta}\sqrt{5 - (\sqrt{5}\sin{\theta})^2}}}

     = \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta}\sqr  t{5 - 5\sin^2{\theta}}}

     = \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta}\sqr  t{5(1 - \sin^2{\theta})}}

     = \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta}\sqr  t{5\cos^2{\theta}}}

     = \int{\frac{\cos{\theta}\,d\theta}{\sqrt{5}\sin{\th  eta}\cos{\theta}}}

     = \int{\frac{d\theta}{\sqrt{5}\sin{\theta}}}

     = -\frac{\log{(\cot{\theta} + \csc{\theta})}}{\sqrt{5}} + C

     = -\frac{\log{\left(\frac{1 + \sqrt{1 - \sin^2{\theta}}}{\sin{\theta}}\right)}}{\sqrt{5}} + C

     = -\frac{\log{\left(\frac{1 + \sqrt{1 - \frac{u^2}{5}}}{\frac{u}{\sqrt{5}}}\right)}}{\sqrt  {5}} + C

     = -\frac{\log{\left(\frac{1 + \sqrt{\frac{5-u^2}{5}}}{\frac{u}{\sqrt{5}}}\right)}}{\sqrt{5}} + C

     = -\frac{\log{\left(\frac{\sqrt{5} + \sqrt{5 - u^2}}{u}\right)}}{\sqrt{5}} + C

     = -\frac{\sqrt{5}\log{\left(\frac{\sqrt{5} + \sqrt{5 - u^2}}{u}\right)}}{5} + C
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