How to solve $\int\frac{e^{-2x}\sin^2x}{1-\sin{2x}}dx$

Question

Consider the integral $$\int\frac{e^{-2x}\sin^2x}{1-\sin{2x}}dx$$

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approch: $$\int\frac{e^{-2x}\sin^2x}{1-\sin{2x}}dx=\int\frac{e^{-2x}\sin^2x}{(\cos x-\sin x)^2}dx=\int \frac{e^{-2x}\sin^2x}{\cos x+\sin x}\times\frac{\cos x+\sin x}{(\cos x-\sin x)^2}dx=$$ $$\int \frac{e^{-2x}\sin^2x}{\cos x+\sin x}d\left(\frac{1}{\cos x-\sin x}\right)$$ and apply integration by parts, there is an easier way?

Any hint would be appreciated.

Answer 1

Let $x=\dfrac\pi4-y,\sin2x=\cos2y,\sin^2x=\dfrac{1-\cos2x}2=\dfrac{1-\sin2y}2$

$$\int\dfrac{e^{-2x}\sin^2x}{1-\sin2x}dx=-\int\dfrac{e^{-2(\pi/4-y)}(1-\sin2y)}{2(1-\cos2y)}dy =-\dfrac1{4e^{\pi/2}}\int e^{2y}\left(\csc^2y-2\cot y\right)dy$$

Now,$$\dfrac{d(e^{2y}\cot y)}{dy}=?$$

Answer 2

What I found nice is $$I=\int\frac{e^{-2x}\sin^2x}{1-\sin{2x}}dx=\int\frac{e^{-2x}\sin^2x}{(\cos x-\sin x)^2}dx$$ Now, cheating a little, assume that $$I=\frac{e^{-2x}(A \cos(x)+B \sin(x))}{\cos(x)-\sin(x)}$$ Differentiate and identify $A$ and $B$.