Integrate $\int{ \left( \frac{1-x}{1+x} \right)^\frac{3}{2}dx}$

Question

Integrate $$\int{ \left(\frac{1-x}{1+x} \right)^\frac{3}{2}dx}$$ I guess that there is sub $x = \cos t$ so integral gets to $$\int{ \left(\tan \frac{t}{2} \right)^3 d\cos t}$$ then I used that $\sin t = \frac{\tan \frac{t}{2}}{(\tan \frac{t}{2})^2 + 1}$ and got $$-2\int {\frac{(sin \frac{t}{2})^4}{(\cos \frac{t}{2})^2}dt}$$but then I stuck with transformations. Please, help.

Answer 1

HINT:

$$\int\left(\frac{1-x}{1+x}\right)^{\frac{3}{2}}\space\text{d}x=$$

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Substitute $u=\frac{1-x}{1+x}$ and $\text{d}u=\left(-\frac{1-x}{(1+x)^2}-\frac{1}{x+1}\right)\space\text{d}x$:

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$$-2\int\frac{u^{\frac{3}{2}}}{(-u-1)^2}\space\text{d}u=$$

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Substitute $s=\sqrt{u}$ and $\text{d}s=\frac{1}{2\sqrt{u}}\space\text{d}u$:

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$$-4\int\frac{s^4}{(-s^2-1)^2}\space\text{d}s=-4\int\frac{s^4}{(s^2+1)^2}\space\text{d}s=-4\int\left[1+\frac{1}{(s^2+1)^2}-\frac{2}{s^2+1}\right]\space\text{d}s$$