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Math Help - Fraction Integral

  1. #1
    Senior Member polymerase's Avatar
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    Fraction Integral

    Find \int\frac{dx}{4-5\sin\:x}

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    Last edited by polymerase; April 15th 2008 at 08:05 AM.
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    Quote Originally Posted by polymerase View Post
    Find \int\frac{dx}{4-5sin\:x}
    Let t = \tan \frac{x}{2} then \sin x =\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}= \frac{1-t^2}{1+t^2} and t' =  \frac{1}{2}\sec^2 \frac{x}{2} = \frac{1}{2}\left( 1+\tan^2 \frac{x}{2}\right) =\frac{1}{2}(1+t^2).

    Thus, it means,
    \int \frac{dx}{4-5\sin x} = \int \frac{dx}{4 - 5 \left( \frac{1-t^2}{1+t^2} \right)} = \int \frac{1+t^2 ~ dx}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{9t^2-1}
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let t = \tan \frac{x}{2} then \sin x =\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}= \frac{1-t^2}{1+t^2} and t' =  \frac{1}{2}\sec^2 \frac{x}{2} = \frac{1}{2}\left( 1+\tan^2 \frac{x}{2}\right) =\frac{1}{2}(1+t^2).

    Thus, it means,
    \int \frac{dx}{4-5\sin x} = \int \frac{dx}{4 - 5 \left( \frac{1-t^2}{1+t^2} \right)} = \int \frac{1+t^2 ~ dx}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{9t^2-1}
    or you could have just let t=tan\frac{x}{2} and then use substitutions of sin\:x=\frac{2t}{1+t^2} and dx=\frac{2}{1+t^2} This leads to \int\frac{dt}{2t^2-5t+2}.
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    Quote Originally Posted by polymerase View Post
    or you could have just let t=tan\frac{x}{2} and then use substitutions of sin\:x=\frac{2t}{1+t^2} and dx=\frac{2}{1+t^2} This leads to \int\frac{dt}{2t^2-5t+2}.
    I made a mistake (I was on the bottom of my vodka bottle when I posted this) \sin x = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}. Now do what I did and you get your answer.
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