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Math Help - Find the real positif number

  1. #1
    Super Member dhiab's Avatar
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    Find the real positif number

    Find the real positif number a such that :
     <br />
\int_{0}^{a}\frac{tan(\frac{\pi }{4}+\frac{x}{2})}{sec^{2}(x)}dx=\frac{1}{16}<br />
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello dhiab
    Quote Originally Posted by dhiab View Post
    Find the real positif number a such that :
     <br />
\int_{0}^{a}\frac{tan(\frac{\pi }{4}+\frac{x}{2})}{sec^{2}(x)}dx=\frac{1}{16}<br />
    There are many positive solutions. Here is the first one. But it doesn't work out very neatly, so check my working!

    \tan\left(\frac{\pi }{4}+\frac{x}{2}\right)=\frac{1+\tan\frac x2}{1-\tan\frac x2}
    =\frac{\cos\frac x2+\sin\frac x2}{\cos\frac x2-\sin\frac x2}

    =\frac{(\cos\frac x2+\sin\frac x2)^2}{\cos^2\frac x2-\sin^2\frac  x2}


    =\frac{1+2\sin\frac x2\cos\frac x2}{\cos x}


    =\frac{1+\sin x}{\cos x}

    \Rightarrow \int_0^a\frac{\tan\left(\frac{\pi }{4}+\frac{x}{2}\right)}{\sec^2x}\;dx = \int_0^a(\cos x + \cos x\sin x)\;dx
    =\int_0^a(\cos x +\tfrac12 \sin2x)\;dx

    =\Big[\sin x -\tfrac14\cos2x\Big]_0^a


     =\sin a - \tfrac14\cos2a +\tfrac14


    =\tfrac1{16}

    \Rightarrow 16\sin a -4(1-2\sin^2a)+4=1

    \Rightarrow 8\sin^2a+16\sin a -1=0


    \Rightarrow \sin a = \frac{-16+\sqrt{256+32}}{16}, taking the positive root
    =\frac{-4+\sqrt{18}}{4}
    So the first positive solution is
    a = \arcsin\left(\frac{-4+\sqrt{18}}{4}\right)
    Grandad
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