Integral of (cosx) / (1+sin^2 x + sin^4 x )
is there a way to do this without the use of a calculator? thanks
Yes, but it's a loo-loo. You can let $\displaystyle u=sin(x), \;\ du=cos(x)dx$
$\displaystyle \int\frac{1}{1+u^{2}+u^{4}}du$
$\displaystyle \int\frac{1}{(u^{2}+u+1)(u^{2}-u+1)}du$
You can use partial fractions. One way. Kriz will probably be along with some show off method
Hello, xfyz!
I had a wild idea . . . but it got me nowhere.
. . (But this trick has worked for me before.)
$\displaystyle \int \frac{\cos x\,dx}{1 + \sin^2\!x + \sin^4\!x}$
Multiply by $\displaystyle \frac{\cos^2\!x}{1-\sin^2\!x}$ . . . . Note that it equals 1.
. . $\displaystyle \frac{\cos x}{1 + \sin^2\!x + \sin^4\!x}\cdot\frac{\cos^2\!x}{1-\sin^2\!x} \;=\;\frac{\cos^3\!x}{1 - \sin^6\!x} $ . . . . very cute. . But now what?
nice. but i don't really see how to take it from here either. i was going to tell you to expand the denominator using the difference of two cubes formula, but then i realized that would just give you back the original function.
hehe, that made me laugh. . . very cute. . But now what?