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Math Help - Inverse Trigonometric Function Integral?

  1. #1
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    Inverse Trigonometric Function Integral?

    If you could help me figure this integral out I would be very grateful.

    top limit: 6
    bottom limit: 4

    of dx / (x-3)sqrt(x^2-6x+8)

    The question involves the arc graphs and whatnot. I'm just really confused.
    Sorry if this is a little messy, I'm not experienced with latex. any help would be appreciated, thanks.


    Im not even sure where to begin. I know i must make a u substitution but I dont know what with.
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    Quote Originally Posted by Torcida1911 View Post
    If you could help me figure this integral out I would be very grateful.

    top limit: 6
    bottom limit: 4

    of dx / (x-3)sqrt(x^2-6x+8)

    The question involves the arc graphs and whatnot. I'm just really confused.
    Sorry if this is a little messy, I'm not experienced with latex. any help would be appreciated, thanks.


    Im not even sure where to begin. I know i must make a u substitution but I dont know what with.
    So this is \int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + 8}}\,dx}

     = \int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + (-3)^2 - (-3)^2 + 8}}\,dx}

     = \int_4^6{\frac{1}{(x - 3)\sqrt{(x - 3)^2 - 1}}\,dx}.


    Now try making the substitution x - 3 = \cosh{t}.
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    Quote Originally Posted by Prove It View Post
    So this is \int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + 8}}\,dx}

     = \int_4^6{\frac{1}{(x - 3)\sqrt{x^2 - 6x + (-3)^2 - (-3)^2 + 8}}\,dx}

     = \int_4^6{\frac{1}{(x - 3)\sqrt{(x - 3)^2 - 1}}\,dx}.


    Now try making the substitution x - 3 = \cosh{t}.
    Alright, I'll try this.

     = \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - 1}}\,dx}.
    =  = \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - (cosht^2 + sinht^2)}}\,dx}.
    =  = \int_4^6{\frac{1}{(cosht)\sqrt{(-sinht^2)}}\,dx}.

    Ah..I've made an error here. I can't square root that..

    Maybe I should set x-3 = u? that way I'll get a similar formula to that of
    d/dx (arcsecx) = 1 / |x|sqrt(x^2 -1) ?


    thank you very much for your answer!!
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  4. #4
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    Quote Originally Posted by Torcida1911 View Post
    Alright, I'll try this.

     = \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - 1}}\,dx}.
    =  = \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - (cosht^2 + sinht^2)}}\,dx}.
    =  = \int_4^6{\frac{1}{(cosht)\sqrt{(-sinht^2)}}\,dx}.

    Ah..I've made an error here. I can't square root that..

    Maybe I should set x-3 = u? that way I'll get a similar formula to that of
    d/dx (arcsecx) = 1 / |x|sqrt(x^2 -1) ?


    thank you very much for your answer!!
    Try setting x-3=\sec(t) \implies dx=\sec(t)\tan(t)dt

    This gives


     \int_4^6{\frac{1}{(x - 3)\sqrt{(x - 3)^2 - 1}}\,dx}=\int \frac{\sec(t)\tan(t)}{\sec(t)\sqrt{\sec^2(t)-1}}dt=\int dt
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  5. #5
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    Quote Originally Posted by Torcida1911 View Post
    If you could help me figure this integral out I would be very grateful.

    top limit: 6
    bottom limit: 4

    of dx / (x-3)sqrt(x^2-6x+8)

    The question involves the arc graphs and whatnot. I'm just really confused.
    Sorry if this is a little messy, I'm not experienced with latex. any help would be appreciated, thanks.


    Im not even sure where to begin. I know i must make a u substitution but I dont know what with.
    They gave you the solution.
    But the problem is they did not notice that the integral is improper.
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    Quote Originally Posted by Torcida1911 View Post
    Alright, I'll try this.

     = \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - 1}}\,dx}.
    =  = \int_4^6{\frac{1}{(cosht)\sqrt{(cosht)^2 - (cosht^2 + sinht^2)}}\,dx}.
    =  = \int_4^6{\frac{1}{(cosht)\sqrt{(-sinht^2)}}\,dx}.

    Ah..I've made an error here. I can't square root that..

    Maybe I should set x-3 = u? that way I'll get a similar formula to that of
    d/dx (arcsecx) = 1 / |x|sqrt(x^2 -1) ?


    thank you very much for your answer!!
    You have forgotten that if x - 3 = \cosh{t} then dx = \sinh{t}\,dt.


    So the integral actually becomes:

    \int{\frac{1}{\cosh{t}\sqrt{\cosh^2{t} - 1}}\,\sinh{t}\,dt}


    Now since \cosh^2{t} - \sinh^2{t} = 1 that means \cosh^2{t} - 1 = \sinh^2{t}.


     = \int{\frac{\sinh{t}}{\cosh{t}\sqrt{\sinh^2{t}}}\,d  t}

     = \int{\frac{\sinh{t}}{\cosh{t}\sinh{t}}\,dt}

     = \int{\frac{1}{\cosh{t}}\,dt}

     = \int{\textrm{sech}\,{t}\,dt}.


    I suggest you consult your tables of integrals to evaluate this integral.




    For an alternative method,

    integral[1/((x - 3)*Sqrt[(x - 3)^2 - 1])] - Wolfram|Alpha

    Click "Show Steps".
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