very basic integral

**Jones** · June 29th 2009, 02:24 AM

Hi,

I have the following integral which i do not know how to solve.

$\int\frac{x^2}{2+x^6}$

Start by using U substitution.

u = $x^3$
du = $3x^2$

now the integral can be seen as $\frac {1}{3} * \int \frac {du}{2+u^2}$

But now what?

**Sean12345** · June 29th 2009, 02:59 AM

Originally Posted by Jones

now the integral can be seen as $\frac {1}{3} * \int \frac {du}{2+u^2}$

But now what?

Hi Jones,

Note that $\int\frac{dx}{a^2+x^2}=\frac{1}{a}\arctan\left(\fr ac{x}{a}\right)$

Does this help?

**Jones** · June 29th 2009, 06:40 AM

Hi,

Yes thank you.

Just one last question.

If you have the Integral $\int \frac {cos x}{4+sin^2x}$

Is it posible to first simplify it down? like so:

$sin^2x = 1-cos^2x$

ending up with $\frac {cosx}{5-cos^2x}$
effectivley canceling out cosx. So that the final Integral becomes $\frac{1}{5-cosx}$

**skeeter** · June 29th 2009, 06:59 AM

Originally Posted by Jones

Hi,

Yes thank you.

Just one last question.

If you have the Integral $\int \frac {cos x}{4+sin^2x}$

Is it posible to first simplify it down? like so:

$sin^2x = 1-cos^2x$

ending up with $\frac {cosx}{5-cos^2x}$
effectivley canceling out cosx. So that the final Integral becomes $\frac{1}{5-cosx}$

same drill with arctan ...

let $u = \sin{x}$

$\int \frac{du}{2^2 + u^2}$

**Krizalid** · June 29th 2009, 07:41 AM

Other way to solve this is to put just one substitution so that the problem can be solved quickly.

If we put $x^3=\sqrt2 t$ the integral becomes $\frac1{3\sqrt2}\int{\frac{dt}{1+t^{2}}}=\frac1{3\s qrt2}\arctan (t)+k,$ and we're done.

----

The key is to turn that into an immediate integral; when we set $u=x^3$ we have a 50% of the work done, and then we need to get our arctagent, so we again put $u=\sqrt2 t,$ and then combine these substitutions to produce $x^3=\sqrt2 t$ to get a faster result. (Apply the same procedure to your other integral.)

**mr fantastic** · June 29th 2009, 02:43 PM

Originally Posted by Jones

Hi,

I have the following integral which i do not know how to solve.

$\int\frac{x^2}{2+x^6} \, {\color{red}dx}$ Mr F says: *Ahem* .... (The main reason why many students unsuccessfully apply the substitution method).

Start by using U substitution.

u = $x^3$
du = $3x^2$

now the integral can be seen as $\frac {1}{3} * \int \frac {du}{2+u^2}$

But now what?

..

**Jones** · June 29th 2009, 03:20 PM

To be honest i never really got the hang of leibniz notation.
Why is that dx so important?

**Jones** · June 30th 2009, 12:33 PM

could you use Taylor series to approximate an integral?

**skeeter** · June 30th 2009, 01:49 PM

Originally Posted by Jones

could you use Taylor series to approximate an integral?

do you have a specific definite integral in mind?

**tom@ballooncalculus** · June 30th 2009, 05:08 PM

Originally Posted by Jones

To be honest i never really got the hang of leibniz notation.
Why is that dx so important?

One thing it's good for is handling substitutions that enable us to work backwards through the chain rule. On the other hand, we often do quite well without it differentiating - and say, e.g.,

$f(g(x))' = f'(g(x)) * g'(x)$

And mapping out the pattern...

... we can often manage equally well without dy/dx when we're travelling 'up' (i.e. integrating)...

Don't integrate - balloontegrate!
Balloon Calculus Forum

**Bruno J.** · June 30th 2009, 05:21 PM

No offence, but these diagrams are extremely confusing to me. I'm not even sure what I'm supposed to see there.

**tom@ballooncalculus** · June 30th 2009, 06:10 PM

None taken. I shouldn't have left out my usual spiel:

"straight continuous lines differentiate downwards (integrate up) with respect to the main variable, the straight dashed line similarly but with respect to the dashed balloon expression."

But I'd be grateful to know if you'd perused some other examples without enlightenment.

Hey - you're expert in the notations you use, so perhaps it's not for you...

Cheers - but take another look!
Tom

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