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Math Help - Trigonometric Substitution Q2

  1. #1
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    Exclamation Trigonometric Substitution Q2

    Question:
     \int\frac{dx}{\sqrt{4x-x^2}}

    How to solve it?
    Please answer in steps , thank you
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int\frac{dx}{\sqrt{4x-x^2}}

    How to solve it?
    Please answer in steps , thank you
    Here, there is no solution with steps.
    we give a "start".
    Try to complete the square for x in 4x-x^2=-(x^2-4x).

    On the other hand, the substitution x=\frac{1}{u} will make the original integral = - \int \frac{du}{u\sqrt{4u-1}} which can be evaluated easily by the substitution z=\sqrt{4u-1}.
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  3. #3
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int\frac{dx}{\sqrt{4x-x^2}}

    How to solve it?
    Please answer in steps , thank you
    Here are some steps to get you started.

    First complete the square on the denominator to get

    4x-x^2=-(x^2-4x)=-(x^2-4x+4)+4=4-(x-2)^2

    From here make the trig sub

    (x-2)=2\sin(y) \implies dx=2\cos(y)dy

    Just sub into the integral and simplify you should end up with

    \int dy
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