integral

• Mar 2nd 2013, 01:44 PM
apatite
integral
what is integral of sqrt(36t^2+9t^4)?
• Mar 2nd 2013, 02:03 PM
Siron
Re: integral
Note that:
$\displaystyle \int \sqrt{36t^2+9t^4}dt = \int \sqrt{9t^2(4+t^2)}dt$

Try to use a good substitution ...
• Mar 2nd 2013, 02:04 PM
Prove It
Re: integral
Quote:

Originally Posted by apatite
what is integral of sqrt(36t^2+9t^4)?

\displaystyle \displaystyle \begin{align*} \sqrt{ 36t^2 + 9t^4 } &= \sqrt{ t^2 \left( 36 + 9t^2 \right) } \\ &= \sqrt{t^2} \sqrt{ 36 + 9t^2 } \\ &= |t| \sqrt{36 + 9t^2} \end{align*}

So if \displaystyle \displaystyle \begin{align*} t \geq 0 \end{align*}

\displaystyle \displaystyle \begin{align*} \int{\sqrt{ 36t^2 + 9t^4 }\,dt} &= \int{ t \, \sqrt{36 + 9t^2}\,dt } \\ &= \frac{1}{18}\int{ 18t \, \sqrt{36 + 9t^2}\,dt } \end{align*}

Now make the substitution \displaystyle \displaystyle \begin{align*} u = 36 + 9t^2 \implies du = 18t\,dt \end{align*} and the integral becomes

\displaystyle \displaystyle \begin{align*} \frac{1}{18}\int{ 18t\, \sqrt{ 36 + 9t^2 } \,dt } &= \frac{1}{18}\int{ \sqrt{u}\,du } \\ &= \frac{1}{18}\int{u^{\frac{1}{2}}\,du} \\ &= \frac{1}{18} \left( 2u^{\frac{3}{2}} \right) + C \\ &= \frac{1}{9}\left( 36 + 9t^2 \right) \sqrt{36 + 9t^2} + C \end{align*}

And if \displaystyle \displaystyle \begin{align*} t < 0 \end{align*} that means \displaystyle \displaystyle \begin{align*} \int{\sqrt{36t^2 + 9t^4}\,dt} = -\frac{1}{9} \left( 36 + 9t^2 \right) \sqrt{ 36 + 9t^2} + C \end{align*}.