Question

Any solution for
Integral of ((1-sin2x)/(1+sin2x))^2?

...
(Please do not suggest sec transform I am not familiar with it.)

Answer 1

Any ideas?

Answer 2

Yes

Answer 3

How to start this integration

Answer 4

We'll go u-sub for 2x,

\[\int\limits \frac{(1-(2 x) \sin )^2}{((2 x) \sin +1)^2} \, dx\]
\[u=2x~~~~ du=2~dx \\ \\ \frac{1}{2}~du=dx\]
\[\frac{1}{2} \int\limits \frac{(1- \sin(u) )^2}{(1+ \sin(u) )^2} \, du\]

Answer 5

I couldn't think of any way besides using sec transform

Answer 6

@UnkleRhaukus @amistre64 @satellite73

Answer 7

please check is square in numerator also or only in denominator.

Answer 8

\[\begin{align*}
\int\left(\frac{1-\sin2x}{1+\sin2x}\right)^2\,\mathrm{d}x&=\int\left(\frac{1-\sin2x}{1+\sin2x}\right)^2\left(\frac{1-\sin2x}{1-\sin2x}\right)^2\,\mathrm{d}x\\[1ex]
&=\int\frac{(1-\sin2x)^4}{(1-\sin^2(2x))^2}\,\mathrm{d}x\\[1ex]
&=\int\frac{(1-\sin2x)^4}{\cos^4(2x)}\,\mathrm{d}x\\[1ex]
&=\int\frac{1-4\sin2x+6\sin^2(2x)-4\sin^3(2x)+\sin^4(2x)}{\cos^4(2x)}\,\mathrm{d}x\\[1ex]
&=\int\sec^4(2x)\,\mathrm{d}x-4\int\tan2x\sec^3(2x)\,\mathrm{d}x\\[1ex]
&\quad\quad +6\int\tan^2(2x)\sec^2(2x)\,\mathrm{d}x-4\int \tan^3(2x)\sec2x\,\mathrm{d}x\\[1ex]
&\quad\quad+\int\tan^4(2x)\,\mathrm{d}x
\end{align*}\]each of which is a standard integral, but I have this nagging feeling there's a more elegant, one-to-two line approach out there somewhere.

Answer 9

Of course, there's always the tangent substitution, but I feel that's even more work than this method.

Answer 10

Thank you .Sam.

Answer 11

Might as well crank that out while I'm here. Let \(y=\tan x\), then \(\sin2x=\dfrac{2y}{1+y^2}\) and \(\mathrm{d}x=\dfrac{\mathrm{d}y}{1+y^2}\), giving
\[\begin{align*}\int\left(\frac{1-\sin2x}{1+\sin2x}\right)^2\,\mathrm{d}x&=\int\left(\frac{1-\frac{2y}{1+y^2}}{1+\frac{2y}{1+y^2}}\right)^2\frac{\mathrm{d}y}{1+y^2}\\[1ex]
&=\int\left(\frac{y-1}{y+1}\right)^4\frac{\mathrm{d}y}{1+y^2}
\end{align*}\]which is doable, but not worth the effort IMO.

Answer 12

I tried to finish with last way but it doesn't work.

Answer 13

It's a perfect candidate for partial fraction decomposition, just a bit tedious.