1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

HINT:

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


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


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


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


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.