1. ## Integrals

Integral of (cosx) / (1+sin^2 x + sin^4 x )

is there a way to do this without the use of a calculator? thanks

2. 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

3. Well sometimes we can turn nasty things into beauty ones, but this one is very nasty.

I haven't developed a quickly way to get this faster, but it looks like the straightforward way can tackle it. (Not nicely, of course.)

4. Weierstrass trick also does not work here well. It leads to a messy integral too.

5. 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?

6. Originally Posted by Soroban
Hello, xfyz!

I had a wild idea . . . but it got me nowhere.
. .
(But this trick has worked for me before.)

Multiply by $\displaystyle \frac{\cos^2\!x}{1-\sin^2\!x}$ . . . . Note that it equals 1.

. . $\displaystyle \frac{\cos x}{1 + \sin^2\!x + \sin^4\!x}\cdot\frac{\cos^2\!x}{1-\sin^2\!x} \;=\;\frac{\cos^3\!x}{1 - \sin^6\!x}$ .

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.
. . . very cute. . But now what?
7. $\displaystyle \frac{\cos^3 x}{1-\sin^6 x} = \frac{(1-\sin^2 x)\cos x}{1-\sin^6 x}$
Let $\displaystyle t = \sin x$ then,
$\displaystyle \int \frac{1-t^2}{1-t^6} dt$