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Math Help - Partial Fraction Integral - Trig with Substitution

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    Partial Fraction Integral - Trig with Substitution

    \displaystyle \int \frac{\cos x dx}{6 \sin^2 x + \sin x}

    let u = \sin x; du = \cos x dx

    \int \frac{du}{6u^2 + u}

    \frac{Ax + B}{6u^2 + u}

    Ax + B = 1

    This leaves me with two variables, but only one equation! What have I missed???

    Thanks!
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    Re: Partial Fraction Integral - Trig with Substitution

    Quote Originally Posted by joatmon View Post
    \displaystyle \int \frac{\cos x dx}{6 \sin^2 x + \sin x}

    let u = \sin x; du = \cos x dx

    \int \frac{du}{6u^2 + u}
    From here \int \frac{du}{6u^2 + u} = \int \frac{du}{u(6u + 1)} , now use some partial fractions?
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  3. #3
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    Re: Partial Fraction Integral - Trig with Substitution

    Quote Originally Posted by joatmon View Post
    \displaystyle \int \frac{\cos x dx}{6 \sin^2 x + \sin x}

    let u = \sin x; du = \cos x dx

    \int \frac{du}{6u^2 + u}

    \frac{Ax + B}{6u^2 + u}

    Ax + B = 1

    This leaves me with two variables, but only one equation! What have I missed???

    Thanks!
    \displaystyle \frac{1}{6u^2 + u} = \frac{1}{u(6u + 1)}

    Then

    \displaystyle \begin{align*} \frac{A}{u} + \frac{B}{6u + 1} &= \frac{1}{u(6u + 1)} \\ \frac{A(6u + 1) + Bu}{u(6u + 1)} &= \frac{1}{u(6u + 1)} \\ A(6u + 1) + Bu &= 1 \\ 6Au + A + Bu &= 1 \\ (6A + B)u + A &= 0u + 1 \\ 6A + B &= 0 \textrm{ and }A = 1 \\ 6 + B &= 0 \\ B &= -6 \end{align*}

    So \displaystyle \frac{1}{u} - \frac{6}{6u + 1} = \frac{1}{u(6u + 1)}.

    Go from here.
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    Re: Partial Fraction Integral - Trig with Substitution

    Thank you! It's been a long day...
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