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.

  • $\begingroup$ Use $\sin^2 A=1-\cos^2 A$, square this and simplify the integrand, so you can integrate yerm by term $\endgroup$ – David Quinn Feb 12 '16 at 20:15



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


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



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