Solving an integral by parts or by substitution?

**UnbeatableZoomy** · February 22nd 2013, 02:36 AM

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

**MINOANMAN** · February 22nd 2013, 03:41 AM

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

**Prove It** · February 22nd 2013, 07:05 AM

Originally Posted by UnbeatableZoomy

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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