1. ## integration2

integral((1-cos^4 x)/(cot x-tan x))

in this i have taken 1 as (sin^2 x+cos^2 x)^2

what should i do after that

2. One way I guess would be to set t=tan x, so you get an integral of a rational expression. Then you would have to solve the resulting integral with partial fractions.
By setting t=tan x I get

$\int \frac{1-\cos^4x}{\cot x - \tan x }\,dx =\int \frac{1-\frac{1}{(1+t^2)^2}}{\frac{1}{t}-t}\frac{\,dt}{1+t^2 }=\int \frac{t^4+2t^2}{\frac{(1-t^2)(1+t^2)^3}{t}}\,dt =\int \frac{t^3(t^2+2)}{(1-t^2)(1+t^2)^3}\,dt$

The rest of it is evaluating this integral. To begin with you should substitute for u=t^2. And so on. Hope this helps a little, and also hope someone has a more elegant solution.

EDIT:

Sorry, that way would have been hard, here is a better one. Note that:
$\frac{1}{\tan x}-\tan x = \frac{1-\tan^2 x}{\tan x}=\frac{2}{\tan 2x}$
$\cos^2 x =\frac{1-\cos 2x}{2}$
So if you use that and substitute $t=\cos 2x$ you should be able to solve the integral np.

3. \displaystyle \begin{align*} \frac{1 - \cos^4{x}}{\cot{x} - \tan{x}} &= \frac{1 - \cos^4{x}}{\frac{\cos{x}}{\sin{x}} - \frac{\sin{x}}{\cos{x}}}\\ &= \frac{1 - \cos^4{x}}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{(1 - \cos^2{x})(1 + \cos^2{x})}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{\sin{x}\cos{x}(1 - \cos^2{x})(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(2 - \sin^2{x})}{1 - 2\sin^2{x}}\end{align*}

So $\displaystyle \int{\frac{1 - \cos^4{x}}{\cot{x} - \tan{x}}\,dx} = \int{\frac{\sin^3{x}(2 - \sin^2{x})\cos{x}}{1 - 2\sin^2{x}}\,dx}$

Make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$ and the integral becomes

$\displaystyle \int{\frac{u^3(2 - u^2)}{1 - 2u^2}\,du}$

Can you go from here?

4. (nevermind, my arithmetic and latex don't mix).

5. Originally Posted by Prove It
\displaystyle \begin{align*} \frac{1 - \cos^4{x}}{\cot{x} - \tan{x}} &= \frac{1 - \cos^4{x}}{\frac{\cos{x}}{\sin{x}} - \frac{\sin{x}}{\cos{x}}}\\ &= \frac{1 - \cos^4{x}}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{(1 - \cos^2{x})(1 + \cos^2{x})}{\frac{\cos^2{x} - \sin^2{x}}{\sin{x}\cos{x}}} \\ &= \frac{\sin{x}\cos{x}(1 - \cos^2{x})(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(1 + \cos^2{x})}{\cos^2{x} - \sin^2{x}} \\ &= \frac{\sin^3{x}\cos{x}(2 - \sin^2{x})}{1 - 2\sin^2{x}}\end{align*}

So $\displaystyle \int{\frac{1 - \cos^4{x}}{\cot{x} - \tan{x}}\,dx} = \int{\frac{\sin^3{x}(2 - \sin^2{x})\cos{x}}{1 - 2\sin^2{x}}\,dx}$

Make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$ and the integral becomes

$\displaystyle \int{\frac{u^3(2 - u^2)}{1 - 2u^2}\,du}$

Can you go from here?
should i divide by u^2

6. No, you should expand the numerator, then apply polynomial long division. It's probably easiest if you write it as $\displaystyle \frac{u^5 - 2u^3}{2u^2 - 1}$.

7. what should i do with the remainder

8. Partial Fractions

9. Hm, sorry for interrupting if you didn't like my solution, but since this is my first post I'd still like for it to be helpful.

As mentioned, using the fact that:
$\tan 2x=\frac{2\tan x}{1-\tan^2 x}$
you can rewrite the denominator as:
$\cot x- \tan x=\frac{1-\tan^2 x}{\tan x}=2\frac{1-\tan^2x}{2\tan x}=\frac{2}{\tan 2x}$.
So with the other substitute for the cosine, you get:
$\int \frac{1-\cos^4x}{\cot x-\tan x}\,dx=\int \frac{1-(\frac{1+\cos 2x}{2})^2}{\frac{2}{\tan 2x}}$
Rewriting:
$\int \frac{1-\frac{1+2\cos 2x + \cos^2 2x}{4}}{2}\tan 2x \,dx=\frac{1}{8}\int (3-2\cos 2x-\cos^2 2x)\frac{\sin 2x}{\cos 2x}\,dx$
So the main idea is that the sine disappears with the substitution
$t=\cos 2x$
$\,dt=-2\sin 2x\,dx$
$\,dx=\frac{\,dt}{-2\sin 2x}$
So returning to the integral:
$\frac{1}{8}\int (3-2t-t^2)\frac{sin 2x}{t} \frac{\,dt}{-2\sin2x}=$
$=\frac{1}{16}\int \frac{t^2+2t-3}{t}\,dt$
Which should prove a fairly easy integral.

Well, anyway... if you're going the other way still consider substituting at least $t=u^2$. Good luck.