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

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

Re: Solving an integral by parts or by substitution?

Quote:

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.

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

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

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