integration2

**prat** · June 4th, 2011, 11:53 PM

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

**Croat** · June 5th, 2011, 01:43 AM

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.

**Prove It** · June 5th, 2011, 02:34 AM

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

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?

**Deveno** · June 5th, 2011, 03:15 AM

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

**prat** · June 5th, 2011, 08:13 AM

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

**Prove It** · June 5th, 2011, 08:19 AM

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

**prat** · June 5th, 2011, 08:26 AM

what should i do with the remainder

**Prove It** · June 5th, 2011, 08:39 AM

Partial Fractions

**Croat** · June 5th, 2011, 08:56 AM

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.

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