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Math Help - Difficult integral!?!

  1. #1
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    Question Difficult integral!?!

    I have no idea where to go on this one:

    using substitution t = sinx evaluate the following integral within the limits pi/2 to pi/6:

    4cosx /(3 + (cosx)^2 dx

    i thought about using t^2 = s(inx)^2 but get in a right mess!

    Can you help?

    Thanks D
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  2. #2
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    Quote Originally Posted by dojo View Post
    I have no idea where to go on this one:

    using substitution t = sinx evaluate the following integral within the limits pi/2 to pi/6:

    4cosx /(3 + (cosx)^2 dx

    i thought about using t^2 = s(inx)^2 but get in a right mess!

    Can you help?

    Thanks D

    Send your question again and this time be careful with the parentheses!

    Tonio
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  3. #3
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    Hello, dojo!

    \int\frac{4\cos x}{3 + \cos^2\!x}\,dx

    We have: . 4\int\frac{\cos x\,dx}{3 + (1-\sin^2\!x)} \;=\;4\int\frac{\cos x\,dx}{4-\sin^2\!x}


    Let: . u \:=\:\sin x \quad\Rightarrow\quad du \:=\:\cos x\,dx


    Substitute: . 4\int\frac{du}{4-u^2} \;=\;4\int\frac{du}{(2+u)(2-u)}


    Apply Partial Fractions and get: . 4\int\bigg[\frac{\frac{1}{4}}{2+u} - \frac{\frac{1}{4}}{2-u}\bigg] du

    . . . . . . =\;\ln|2+u| - \ln|2-u| + C \;=\; \ln\left|\frac{2+u}{2-u}\right| + C


    Back-substitute: . \ln\left|\frac{2+\sin x}{2 - \sin x}\right| + C

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