Question

integral of x^2tanx/(1+cos^4x)

Answer 1

I'm haven't' seen if it works but you can try to see if works
let u=tan(x/2)

Answer 2

that might be too ugly still

Answer 3

mostly because that x term i think

Answer 4

there is another weird trig substitution to try but i can't remember it

Answer 5

looks hard for wolfram too
http://www.wolframalpha.com/input/?i=%5Cint+x%5E2tanx%2F%281%2Bcos%5E4x%29

Answer 6

maybe there is some kinda neat trick

Answer 7

any hint @noushinrahdar given?

Answer 8

actually I think I got it...

integration by parts will be needed

Answer 9

\[\int\limits_{}^{}x^2 \cdot \frac{\tan(x)}{1+\cos^4(x) }dx \]
x^2 will be the one you are differentiating
the other function you will need to integrate
i preferred writing tan(x)/(1+cos^4(x)) in terms of sin and cos (of course you need to get rid of the compound fraction)

Answer 10

With IBP, set
\[u=x^2~~\implies~~du=2x~dx\\
dv=\frac{\tan x}{1+\cos^4x}~dx\]
Integrating for \(v\):
\[\begin{align*}
v&=\int\frac{\tan x}{1+\cos^4x}~dx&t=\cos x\\\\
&=-\int\frac{dt}{t(1+t^4)}~dt\\\\
&=-\int\frac{A}{t}~dt-\int\frac{Bt^3+Ct^2+Dt+E}{1+t^4}~dt
\end{align*}\]
\[
-1=A+At^4+Bt^4+Ct^3+Dt^2+Et~~\iff~~\begin{cases}A=-1\\A+B=0\\C=D=E=0\end{cases}\]
\[\begin{align*}
v&=\int\frac{dt}{t}-\int\frac{t^3}{1+t^4}~dt\\\\
&=\ln|t|-\frac{1}{4}\int\frac{ds}{s}&s=1+t^4\\\\
&=\ln|t|-\frac{1}{4}\ln|s|\\\\
&=\ln|\cos x|-\frac{1}{4}\ln(1+\cos^4x)
\end{align*}\]
So you have
\[\int\frac{x^2\tan x}{1+\cos^4x}~dx=\\
x^2\left(\ln|\cos x|-\frac{1}{4}\ln(1+\cos^4x)\right)-2\int x\left(\ln|\cos x|-\frac{1}{4}\ln(1+\cos^4x)\right)~dx\]
Doesn't look elementary...

Answer 11

Is this a definite integral? @noushinrahdar

Answer 12

@SithsAndGiggles I think it is an elementary integral. I was able to integrated suggesting that way I mentioned.

Answer 13

at least I think I did finish it
let me see if i can do it again

Answer 14

brb

Answer 15

Take your time, I'm trying to see if Mathematica can spit out a result.

Answer 16

\[\int\limits_{}^{}x^2 \cdot \frac{\tan(x)}{1+\cos^4(x)} dx\]
So first I suggested writing in terms of sin and cos
\[\int\limits_{}^{}x^2 \cdot \frac{\sin(x)}{\cos(x)+\cos^5(x)} dx \]
Then I perform integration by parts but I took that one function to the side to find the antiderivative first of course since I didn't know it off the bat.
\[\int\limits_{}^{}\frac{\sin(x)}{\cos(x)+\cos^5(x)} dx \text{ \let } u =\cos(x) => du=-\sin(x) dx \\ \int\limits_{}^{}\frac{-du}{u+u^5} du = \int\limits_{}^{} \frac{-du}{u^5(\frac{1}{u^4}+1)} \text{ \let } b=\frac{1}{u^4}+1=> \\ db=-\frac{4}{u^5} du => \frac{db}{4} =\frac{-du}{u^5} \\ \int\limits_{}^{} \frac{db}{4b}=\frac{1}{4} \ln|b|+C=\frac{1}{4} \ln|\frac{1}{u^4}+1|+C =\frac{1}{4} \ln|\frac{1}{\cos^4(x)}+1|+C\]

Ok so going back to the integration by parts idea:

\[\int\limits_{}^{}x^2 \frac{\sin(x)}{\cos(x)+\cos^5(x)} dx \\ =x^2 \cdot \frac{1}{4} \ln |\frac{1}{\cos^4(x)}+1| - \int\limits_{}^{}2x \cdot \frac{1}{4} \ln|\frac{1}{\cos^4(x)}+1| dx \]

Now we need to perform integration by parts again. Trying to remember everything I did...
So lets look at integrating that ln part \[\int\limits_{}^{} \ln |\frac{1}{\cos^4(x)}+1| dx =x \ln|\frac{1}{\cos^4(x)}+1|-\int\limits_{}^{} x \cdot 4 \frac{\sec^4(x) \tan(x)}{\frac{1}{cos^4(x)}+1}dx\]
\[=x \ln|\frac{1}{\cos^4(x)}+1|-\int\limits_{}^{}4x \frac{\sin(x)}{\cos(x)+\cos^5(x)} dx \]

maybe I lied

Answer 17

hey @SithsAndGiggles can mathematica integrate ln|sec^4(x)+1|
wolfram can't

(I think you were right. This is definitely not an elementary integral. I think I was sleepy when I tried to find away to integrate the above. :( )

Answer 18

http://openstudy.com/updates/512c480de4b02acc415db053
I think it is possible you were right
this was maybe meant to be an definite integral

Answer 19

With the absolute value, no. It just returns the input. Without, I'm getting this non-elementary expression with something called a polylogarithm and a bunch of elementary functions in terms of imaginary numbers