Question

Интеграл dx/ (sin^2(x)*(1-cos(x)))

Answer 1

$\large \int\frac{dx}{sin^2x(1-cosx)}=\int\frac{2(1+t^2)^3}{8(1+t^2)t^4}dt=\frac{1}{4}\int\frac{(1+t^2)^2}{t^4}dt=\\\\=\frac{1}{4}\int(1+\frac{2}{t^2}+\frac{1}{t^4})dt=\frac{t}{4}-\frac{1}{2t}-\frac{1}{12t^3}+C=\\\\=\frac{tg\frac{x}{2}}{4}-\frac{ctg\frac{x}{2}}{2}-\frac{ctg^3\frac{x}{2}}{12}+C\\\\\\t=tg\frac{x}{2}\\x=2arctgt\\dx=\frac{2dt}{1+t^2}\\sinx=\frac{2t}{1+t^2}\ ;cosx=\frac{1-t^2}{1+t^2}$

Answer 2

$\displaystyle\int\frac{dx}{\sin^2x\cdot(1-\cos x)}=\int\frac{(1+\cos x)\,dx}{\sin^2 x\cdot(1-\cos^2x)}=\int \frac{dx}{\sin^4 x}+\int\frac{\cos x\,dx}{\sin^4 x}$

Первый интеграл (+C опущено):
$\displaystyle\int \frac{dx}{\sin^4 x}=\int\frac1{\sin^2x}\cdot\frac{dx}{\sin^2 x}=-\int(\mathrm{ctg}^2 x+1)\, d\mathop{\mathrm{ctg}} x=-\frac{\mathrm{ctg}^3 x}3-\mathop{\mathrm{ctg}} x$

Второй интеграл (+С опущено):
$\displaystyle\int\frac{\cos x\,dx}{\sin^4 x}=\int\frac{d\sin x}{\sin^4 x}=-\frac1{3\sin^3 x}$

Итого: 
$\displaystyle\int\frac{dx}{\sin^2x\cdot(1-\cos x)}=C-\frac{\mathrm{ctg}^3 x}3-\mathop{\mathrm{ctg}} x-\frac1{3\sin^3 x}$