# Solving an integral by parts or by substitution?

• Feb 22nd 2013, 03:36 AM
UnbeatableZoomy
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
• Feb 22nd 2013, 04:41 AM
MINOANMAN
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
• Feb 22nd 2013, 08:05 AM
Prove It
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 \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*}