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Math Help - Solving an integral by parts or by substitution?

  1. #1
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    Solving an integral by parts or by substitution?

    Hi all,

    I'm wondering is it bettetr to solve the integral v/((1 + v^2)^1/2) dv using integration by parts, making v'=((1+v^2)^-1/2) or to solve it using substitution, making v=sinh(x)?

    It can probably be solved both ways though I'm wondering which way is generally an easier method to use.

    Thanks
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    Re: Solving an integral by parts or by substitution?

    Multiply both (num +denom.) by 2 to obtain[ 2v/2(1+v^2)^1/2]dv then integrate directly to obtain sqrroot(1+v^2) +C
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    Re: Solving an integral by parts or by substitution?

    Quote Originally Posted by UnbeatableZoomy View Post
    Hi all,

    I'm wondering is it bettetr to solve the integral v/((1 + v^2)^1/2) dv using integration by parts, making v'=((1+v^2)^-1/2) or to solve it using substitution, making v=sinh(x)?

    It can probably be solved both ways though I'm wondering which way is generally an easier method to use.

    Thanks
    To answer your question, neither of your methods is the most efficient. Always look for SIMPLE substitutions.

    \displaystyle \begin{align*} \int{\frac{v}{\left( 1 + v^2 \right)^{\frac{1}{2}} } \,dv} &= \frac{1}{2} \int{\frac{2v}{ \left( 1 + v^2 \right)^{\frac{1}{2}} }\,dv} \end{align*}

    Now let \displaystyle \begin{align*} u = 1 + v^2 \implies du = 2v\,dv \end{align*} and the integral becomes

    \displaystyle \begin{align*} \frac{1}{2} \int{\frac{2v}{\left( 1 + v^2 \right) ^{\frac{1}{2}} }\,dv} &= \frac{1}{2} \int{ \frac{1}{u^{\frac{1}{2}}}\,du } \\ &= \frac{1}{2} \int{u^{-\frac{1}{2}}\,du} \\ &= \frac{1}{2} \left( 2u^{\frac{1}{2}} \right) + C \\ &= u^{\frac{1}{2}} + C \\ &= \left( 1 + v^2 \right)^{\frac{1}{2}} + C \end{align*}
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