1. ## very basic integral

Hi,

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

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

Start by using U substitution.

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

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

But now what?

2. Originally Posted by Jones

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

But now what?
Hi Jones,

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

Does this help?

3. Hi,

Yes thank you.

Just one last question.

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

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

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

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

4. Originally Posted by Jones
Hi,

Yes thank you.

Just one last question.

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

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

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

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

let $\displaystyle u = \sin{x}$

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

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

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

----

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

6. Originally Posted by Jones
Hi,

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

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

Start by using U substitution.

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

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

But now what?
..

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

8. could you use Taylor series to approximate an integral?

9. Originally Posted by Jones
could you use Taylor series to approximate an integral?
do you have a specific definite integral in mind?

10. 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.,

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

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

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

12. 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