Integral Indefine

**dianjoey** · April 28th 2009, 03:07 PM

$\int{x(2-x^2)^3}\,dx$

How could I get the -1/2???

**Danny** · April 28th 2009, 03:25 PM

Originally Posted by dianjoey

$\int{x(2-x^2)^3}\,dx$

How could I get the -1/2???

I think what you want is
$<br /> \int x(2-x^2)^3\,dx = -\frac{1}{2}\int (2-x^2)^3\, d(2-x^2)<br />$

but it's really better if you use a substitution. Let $u = 2-x^2$ (the hard part of this). Then $du = -2 x \,dx$ so $x\,dx = -\frac{1}{2}\,du$ so the original problem becomes

$<br /> -\frac{1}{2} \int u^3\,du<br />$

**dianjoey** · April 28th 2009, 03:37 PM

Hi Danny,

I was asking where the $-\frac 12\$ come from...
I don't know how to get that...

Thread: Integral Indefine

LinkBack

Thread Tools

Display

Integral Indefine

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Explaining result: comparing line integral and two-form integral

[SOLVED] Line integral, Cauchy's integral formula

write [integral of] dx/SQRT(sinx) in terms of an elliptic integral

Search Tags